SAT Math Practice Test: Algebra and Functions section with four sample questions.

Let’s solve each problem one by one, carefully checking our work.

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Problem 1: Which of the following expressions is equal to 64!?

We know that factorial means multiplying all positive integers up to that number. So:

64! = 64 × 63 × 62 × ... × 2 × 1

Now look at the options:

A. 2·3! → 2×6 = 12 → too small
B. 4! → 24 → too small
C. 8! → 40320 → still way smaller than 64!
D. 8!! → This is “double factorial” — for even numbers, it means multiply every other number down to 2.

So 8!! = 8 × 6 × 4 × 2 = 384 → still tiny compared to 64!

Wait — none of these seem right? But maybe we’re missing something.

Actually, let’s think differently. Is there a trick?

Note: 64 = 8², but that doesn’t help directly.

Wait — perhaps the question has a typo or misprint? Or maybe we’re supposed to recognize that none are equal? But that’s not an option.

Hold on — let’s check if any expression could be interpreted as 64!.

Option D: 8!! — double factorial — no, that’s 384.

But wait — what if “8!!” was meant to be “(8!)!”? That would be huge — 40320! — which is WAY bigger than 64!.

Nope.

Alternatively, maybe the question is asking which is *equal* to 64! in value? None are.

But let’s re-read: “Which of the following expressions is equal to 64!?”

Perhaps it’s a trick question? Or maybe I’m overcomplicating.

Wait — actually, 64! is just 64!, and none of the options equal that. But since this is a multiple choice test, maybe there’s a mistake in my reasoning.

Let me calculate 8! again: 8×7×6×5×4×3×2×1 = 40320

64! is astronomically larger.

Unless... is “8!!” meaning something else? In some contexts, !! can mean iterated factorial, but that’s non-standard.

Actually, standard double factorial for even n: n!! = n×(n-2)×...×2

So 8!! = 8×6×4×2 = 384

Still not close.

Wait — perhaps the question is miswritten, and it’s supposed to be “which is equal to 8!”? Then C would be correct.

But it says 64!.

Alternatively, maybe it’s a notation issue — sometimes people write n! for factorial, but here it’s clear.

I think there might be an error in the problem, but since we have to choose, and none match, perhaps the intended answer is D, assuming they meant something else? No, that’s guessing.

Wait — let’s look at Problem 2 — maybe it’s related? No.

Another thought: 64 = 2^6, but that doesn’t help with factorial.

Perhaps the question is “which is equal to 64” not “64!”? Let me check the image description — it says “64!” with exclamation mark, so factorial.

But in the user’s text, it’s written as “64!” — so factorial.

Given that, and since none of the options are close, I suspect a typo, and perhaps it’s supposed to be “8!” — then C is correct.

Maybe “64!” is a misprint for “8!”.

In many tests, such typos happen.

Alternatively, let’s see option D: 8!! — if it were (8!)!, that’s too big.

Or if it’s 8^2 = 64, but that’s not factorial.

I think the most reasonable assumption is that the question meant “which is equal to 8!” — then C. 8! is correct.

Because 8! = 40320, and 64! is much larger, but perhaps in context, it’s a mistake.

Maybe “64!” is meant to be “the factorial of 8 squared” but 8^2=64, same thing.

I recall that sometimes in older texts, ! might mean something else, but unlikely.

Another idea: perhaps “64!” is not factorial, but exclamation for emphasis? But that doesn’t make sense in math context.

Looking back at the problem list, Problem 2 has factorials, so likely it is factorial.

Perhaps the answer is none, but since it’s multiple choice, and D is 8!!, which is not standard, maybe it’s a distractor.

Let’s move to Problem 2 and come back.

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Problem 2: If x ≠ 1, which of the following is equivalent to [(x+1)/(x-1)]² ?

We need to simplify or find equivalent expression.

[(x+1)/(x-1)]² = (x+1)² / (x-1)²

Now look at options:

A. [(1-x²)/(1+x)]² — let’s simplify inside: (1-x²)/(1+x) = [(1-x)(1+x)]/(1+x) = 1-x, for x≠-1

So [(1-x)]² = (1-x)²

But we have (x+1)²/(x-1)² = [(x+1)/(x-1)]²

Note that (x-1)² = (1-x)², since squaring removes sign.

And (x+1)² = (1+x)²

So [(x+1)/(x-1)]² = [(x+1)²] / [(x-1)²] = [(1+x)²] / [(1-x)²] because (x-1)² = (1-x)²

Now, [(1+x)/(1-x)]²

But option A is [(1-x²)/(1+x)]² = [1-x]² as above, which is (1-x)², not the same.

Option B: [(1-x²)/(1+x)]² — same as A? Wait, in the image, A and B look similar.

In the user’s text:

A. \left(\frac{1 - x^{2}}{1 + x}\right)^{2}

B. \left[\frac{(1 - x^{2})}{(1 + x)}\right]^{2} — same as A? Probably typo in transcription.

In original image, likely different.

Assuming standard, let’s compute numerically.

Pick x=2.

Then [(2+1)/(2-1)]² = (3/1)² = 9

Now check options:

A. [(1-4)/(1+2)]² = [(-3)/3]² = (-1)² = 1 ≠9

B. same as A? Or perhaps B is different.

In user’s text, B is written same as A, but probably it’s different.

Looking back: "A. \left(\frac{1 - x^{2}}{1 + x}\right)^{2}" and "B. \left[\frac{(1 - x^{2})}{(1 + x)}\right]^{2}" — identical.

That can’t be. Perhaps in image, B is different.

Maybe B is \left[\frac{1 - x}{1 + x}\right]^2 or something.

Another option: C. (x-1)^{-2} = 1/(x-1)^2

At x=2, 1/(1)^2=1≠9

D. (1+x)^2 — at x=2, 9 — oh! (1+2)^2=9, which matches.

Is [(x+1)/(x-1)]² equal to (1+x)^2? Only if denominator is 1, which it’s not.

At x=2, (1+x)^2=9, and original is 9, but is it always true?

Original: [(x+1)/(x-1)]²

At x=3: [(4)/(2)]²=4

(1+x)^2=16≠4

So not D.

At x=2, D gave 9, same as original, but at x=3, original is 4, D is 16, not equal.

So not D.

What about C: (x-1)^{-2} = 1/(x-1)^2, at x=2, 1/1=1≠9

None seem to work.

Perhaps I miscalculated.

Original at x=2: (3/1)^2=9

Let me try option A: [(1-x^2)/(1+x)]^2 = [(1-4)/(3)]^2 = [-3/3]^2=1

Not 9.

Option B same.

Option C: (x-1)^{-2} = 1/(x-1)^2 =1/1=1

Option D: (1+x)^2=9 at x=2, but as above, not generally.

Unless the expression is different.

Another thought: perhaps "equivalent" means algebraically equivalent after simplification.

[(x+1)/(x-1)]^2

Can we write it as something else?

Note that (x+1)/(x-1) = [ (x-1) +2 ] / (x-1) = 1 + 2/(x-1), but squaring that is messy.

Perhaps rationalize or something.

Let's expand both numerator and denominator.

Numerator: (x+1)^2 = x^2 +2x+1

Denominator: (x-1)^2 = x^2 -2x+1

So ratio is (x^2+2x+1)/(x^2-2x+1)

Now look at options.

Option A: [(1-x^2)/(1+x)]^2 = [-(x^2-1)/(x+1)]^2 = [-(x-1)(x+1)/(x+1)]^2 = [-(x-1)]^2 = (x-1)^2

So (x-1)^2

But we have (x^2+2x+1)/(x^2-2x+1) = (x+1)^2/(x-1)^2

Which is not (x-1)^2.

Unless for specific x, but not generally.

Perhaps the question is to simplify, and one of the options is equal.

Let's try x=0.

Original: [(0+1)/(0-1)]^2 = (1/-1)^2 =1

Option A: [(1-0)/(1+0)]^2 = (1/1)^2=1 — oh! At x=0, both are 1.

At x=2, original is 9, option A is [(1-4)/(1+2)]^2 = [-3/3]^2=1, not 9. Contradiction.

At x=0, original is (1/-1)^2=1, option A is (1/1)^2=1, good.

At x=2, original (3/1)^2=9, option A ( -3/3)^2=1, not 9.

So not equal.

Unless I have the wrong option.

Perhaps B is different. In the user's text, B is written the same as A, but maybe in image it's \left[\frac{1 - x}{1 + x}\right]^2 or something.

Let me assume that B is \left(\frac{1 - x}{1 + x}\right)^2

At x=2, (1-2)/(1+2) = (-1)/3, squared is 1/9 ≠9

Not good.

C: (x-1)^{-2} = 1/(x-1)^2, at x=2, 1/1=1≠9

D: (1+x)^2=9 at x=2, but at x=0, (1+0)^2=1, original is 1, good; at x=3, original (4/2)^2=4, D (4)^2=16≠4.

So not.

Perhaps the expression is [(x+1)/(x-1)]^2, and we need to see which is identical.

Another idea: perhaps "equivalent" means has the same value for all x, so must be identical function.

Let me set y = [(x+1)/(x-1)]^2

Can I write it as k * something.

Notice that (x+1)/(x-1) = - (x+1)/(1-x)

So [(x+1)/(x-1)]^2 = [ (x+1)/(1-x) ]^2 since square.

And (x+1)/(1-x) = - (x+1)/(x-1), but squared, same as before.

(x+1)/(1-x) = [ -(1-x) +2 ] / (1-x) = -1 + 2/(1-x), not helpful.

Let's look at option A again: [(1-x^2)/(1+x)]^2 = [ (1-x)(1+x)/(1+x) ]^2 = (1-x)^2 for x≠-1

So (1-x)^2

But our expression is [(x+1)/(x-1)]^2 = (x+1)^2 / (x-1)^2 = (x+1)^2 / (1-x)^2 since (x-1)^2 = (1-x)^2

So = [ (x+1)/(1-x) ]^2

Whereas option A is (1-x)^2

So not the same.

Unless they have [ (x+1)/(1-x) ]^2, but that's not listed.

Perhaps in the options, there is \left( \frac{x+1}{1-x} \right)^2, but in user's text, it's not.

Let's read the user's input carefully:

"2) If x ≠ 1, which of the following is equivalent to \left(\frac{x+1}{x-1}\right)^{2} ?"

Options:

"A. \left(\frac{1 - x^{2}}{1 + x}\right)^{2}"

"B. \left[\frac{(1 - x^{2})}{(1 + x)}\right]^{2}" — same as A

"C. (x - 1)^{-2}"

"D. (1 + x)^{2}"

All seem incorrect based on calculation.

But at x=0, original is 1, A is 1, C is ( -1)^{-2} = 1, D is 1, so all give 1 at x=0.

At x=2, original 9, A: [(1-4)/(1+2)]^2 = [-3/3]^2=1, C: (2-1)^{-2}=1, D: (1+2)^2=9 — so D gives 9, same as original.

At x=3, original [(4)/(2)]^2=4, D: (1+3)^2=16≠4

So not D.

At x=0.5, original [(1.5)/(-0.5)]^2 = (-3)^2=9

D: (1+0.5)^2=2.25≠9

So not.

Perhaps for x>1 or something, but no.

Another thought: perhaps "equivalent" means can be simplified to that, but none match.

Let's calculate the expression: \left(\frac{x+1}{x-1}\right)^2 = \frac{(x+1)^2}{(x-1)^2}

Now, is this equal to any option?

Option C: (x-1)^{-2} = 1/(x-1)^2, which is different.

Unless they have \frac{(x+1)^2}{(x-1)^2}, but not listed.

Perhaps in option B, it's different. In some fonts, it might be \left( \frac{1 - x}{1 + x} \right)^2, but that's not what's written.

Let's assume that B is \left( \frac{1 - x}{1 + x} \right)^2

At x=2, (1-2)/(1+2) = -1/3, squared 1/9 ≠9

Not.

Or \left( \frac{x-1}{x+1} \right)^2, at x=2, (1/3)^2=1/9≠9

Not.

Perhaps the answer is not among, but that can't be.

Let's try to see if there's a identity.

Notice that \frac{x+1}{x-1} = \frac{(x-1) +2}{x-1} = 1 + \frac{2}{x-1}

So [1 + 2/(x-1)]^2 = 1 + 4/(x-1) + 4/(x-1)^2

Not matching any.

Perhaps for the purpose of this test, they want us to recognize that (x-1)^2 = (1-x)^2, but still.

Let's look at option A: \left(\frac{1 - x^{2}}{1 + x}\right)^{2} = \left( \frac{(1-x)(1+x)}{1+x} \right)^2 = (1-x)^2 for x≠-1

And our expression is \left( \frac{x+1}{x-1} \right)^2 = \left( \frac{x+1}{-(1-x)} \right)^2 = \left( - \frac{x+1}{1-x} \right)^2 = \left( \frac{x+1}{1-x} \right)^2

So \left( \frac{x+1}{1-x} \right)^2 vs (1-x)^2

Clearly different.

Unless they have \left( \frac{1+x}{1-x} \right)^2, which is the same as ours, since (x+1)=(1+x), and (x-1)=-(1-x), but squared, so same as \left( \frac{1+x}{1-x} \right)^2

Oh! So \left( \frac{x+1}{x-1} \right)^2 = \left( \frac{1+x}{-(1-x)} \right)^2 = \left( - \frac{1+x}{1-x} \right)^2 = \left( \frac{1+x}{1-x} \right)^2

So it is equal to \left( \frac{1+x}{1-x} \right)^2

Now, is that in the options? Not explicitly, but let's see if any option matches this.

Option A is \left( \frac{1 - x^{2}}{1 + x} \right)^2 = (1-x)^2 as above.

Not the same.

Perhaps in the list, there is an option like that, but in user's text, it's not.

Maybe for problem 2, the correct choice is not listed, but that can't be.

Another idea: perhaps "equivalent" means has the same domain or something, but unlikely.

Let's calculate the difference.

Set f(x) = [(x+1)/(x-1)]^2

g(x) = (1+x)^2

As above, not equal.

Perhaps they mean which is equal when simplified, but all are already simple.

Let's try x= -2.

Original: [(-2+1)/(-2-1)]^2 = [ (-1)/(-3) ]^2 = (1/3)^2 = 1/9

Option A: [(1-4)/(1-2)]^2 = [ (-3)/(-1) ]^2 = 3^2 =9 ≠1/9

Option C: ( -2-1)^{-2} = (-3)^{-2} = 1/9 — oh! At x= -2, C gives 1/9, same as original.

At x=2, original 9, C: (2-1)^{-2} =1^{-2}=1≠9

At x=0, original 1, C: (0-1)^{-2} = (-1)^{-2} =1, good.

At x=3, original (4/2)^2=4, C: (3-1)^{-2} =2^{-2}=1/4≠4

So not.

At x= -2, both 1/9, but not generally.

Perhaps for x<1 or something.

I think there might be a mistake in the problem or my understanding.

Let's move to Problem 3.

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Problem 3: If 2x + y = 2, what is the value of \frac{16^x}{4^y} ?

First, express with same base.

16 = 4^2, so 16^x = (4^2)^x = 4^{2x}

So \frac{16^x}{4^y} = \frac{4^{2x}}{4^y} = 4^{2x - y}

From the equation: 2x + y = 2

We need 2x - y.

Let me denote a = 2x, b = y, so a + b = 2, and we want 4^{a - b}

But I have a + b = 2, and I need a - b.

I have two variables, one equation, so I need another relation, but there isn't.

Unless I can find a - b from a + b = 2, but I can't without more information.

For example, if x=1, y=0, then 2(1)+0=2, good, then 16^1 / 4^0 = 16/1 = 16

If x=0, y=2, then 2(0)+2=2, good, then 16^0 / 4^2 = 1/16

Different values! 16 vs 1/16, so not constant? But the question asks for "the value", implying it's unique.

That can't be. Did I make a mistake?

16^x / 4^y = 4^{2x} / 4^y = 4^{2x - y}

From 2x + y = 2, I can solve for y: y = 2 - 2x

Then 2x - y = 2x - (2 - 2x) = 2x -2 +2x = 4x -2

So 4^{4x -2}

Which depends on x, so not constant.

But the question implies it's a specific value, so perhaps I misread.

The expression is \frac{16^x}{4^y}, and 2x + y = 2.

But as above, it depends on x.

Unless y is in the exponent differently.

Perhaps it's 16^x divided by 4^y, yes.

Another thought: perhaps "16^x" and "4^y" , and 16=2^4, 4=2^2, so (2^4)^x / (2^2)^y = 2^{4x} / 2^{2y} = 2^{4x - 2y}

From 2x + y = 2, multiply by 2: 4x + 2y = 4

But I have 4x - 2y, which is different.

Let me set u = 4x - 2y

From 2x + y = 2, I can't determine u uniquely.

For example, if x=1, y=0, u=4-0=4, 2^4=16

If x=0, y=2, u=0-4= -4, 2^{-4}=1/16

Same as before.

But the options are A.4 B.8 C.16 D.32, all positive, and 1/16 is not there, so perhaps only certain values, but the equation allows multiple.

Unless there's a constraint I missed.

Perhaps "2x + y = 2" is for the exponents, but no.

Another idea: perhaps the expression is \frac{16^x}{4^y} and they want it in terms of the equation, but still.

Let's look at the expression: \frac{16^x}{4^y} = (4^2)^x / 4^y = 4^{2x} / 4^y = 4^{2x - y}

And from 2x + y = 2, let me add the two equations.

Let S = 2x + y = 2

D = 2x - y = ?

Then S + D = 4x, S - D = 2y, not helpful.

From S = 2x + y = 2, and I want D = 2x - y.

Then D = 2x - y = (2x + y) - 2y = 2 - 2y

Or D = 2x - y = 2(2x + y) - 3y - 2x, messy.

D = 2x - y = a, then from S=2x+y=2, adding: 2a = 4x, so x = a/2, then from S, 2*(a/2) + y =2, so a + y =2, y=2-a

But no new info.

Perhaps the problem is to express it, but the options are numbers, so must be constant.

Unless I miscalculated the base.

16^x / 4^y = (2^4)^x / (2^2)^y = 2^{4x} / 2^{2y} = 2^{4x - 2y}

Now, 4x - 2y = 2(2x - y)

From 2x + y = 2, let me solve for 2x - y.

Let me set u = 2x - y

Then I have:

2x + y = 2

2x - y = u

Add them: 4x = 2 + u, so x = (2+u)/4

Subtract: 2y = 2 - u, y = (2-u)/2

But no constraint, so u can be anything.

For example, if u=4, x= (2+4)/4=1.5, y=(2-4)/2= -1, then 2x+y=3-1=2, good, then 2^{4*1.5 - 2*(-1)} = 2^{6 +2} =2^8=256, not in options.

Earlier with x=1,y=0, 2^{4-0}=16

With x=0,y=2, 2^{0-4}=2^{-4}=1/16

So varies.

But perhaps in the context, x and y are integers or something, but not specified.

Maybe the expression is \frac{16^x}{4^y} and they mean 16 to the x, etc, but same.

Another possibility: perhaps "2x + y = 2" is not the constraint, but part of the expression, but no, it's given as "if 2x + y = 2".

Perhaps it's 2^{x} + y = 2, but that would be different.

Let's read the user's input: "3) If 2x + y = 2, what is the value of \frac{16^{x}}{4^{y}} ?"

Yes.

Perhaps it's \frac{16^x}{4^y} = 4^{2x} / 4^y = 4^{2x - y}, and from 2x + y = 2, if I let z = 2x - y, then as above.

Notice that (2x + y) + (2x - y) = 4x, (2x + y) - (2x - y) = 2y, not helping.

Let me calculate 2x - y in terms of the given.

From 2x + y = 2, y = 2 - 2x

Then 2x - y = 2x - (2 - 2x) = 4x - 2

So 4^{4x - 2} = (2^2)^{4x - 2} = 2^{8x - 4}

Still depends on x.

Unless the problem has a typo, and it's 2x - y = 2 or something.

Perhaps it's 2^x + 2^y = 2, but that would be different.

Another idea: perhaps "2x + y = 2" is for the exponents in a different way, but unlikely.

Let's look at the options: 4,8,16,32, all powers of 2.

From earlier, when x=1, y=0, we get 16, which is option C.

When x=0.5, y=1, then 2*0.5 +1=1+1=2, good, then 16^{0.5} / 4^1 = 4 / 4 =1, not in options.

When x=1.5, y=2-3= -1, then 16^{1.5} / 4^{-1} = (64) / (1/4) = 64 * 4 = 256, not in options.

So only when y=0, x=1, we get 16, which is in options.

Perhaps they assume y=0, but not stated.

Maybe the expression is \frac{16^x}{4^y} and they want it for the case where it's integer or something.

Perhaps "2x + y = 2" and x,y real, but the value is not constant, so maybe the problem is to find possible values, but it says "the value".

Another thought: perhaps it's \frac{16^x}{4^y} = 4^{2x - y}, and from 2x + y = 2, if I consider that 2x - y = k, then as above, but perhaps in the context of the test, they expect us to use the equation to substitute.

Let me set t = x, then y = 2 - 2t, then expression = 4^{2t - (2-2t)} = 4^{2t -2 +2t} = 4^{4t -2} = (2^2)^{4t-2} = 2^{8t -4}

To make this constant, it must be that 8t-4 is constant, which it's not.

Unless t is fixed, but not.

Perhaps the problem is "if 2x + y = 2, and x,y are such that the expression is defined, find its value" but still not unique.

Let's check the next problem; perhaps I can come back.

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Problem 4: If |7^{z+1}| = 4, what is the value of z?

Absolute value of 7^{z+1} = 4.

But 7^{z+1} is always positive for real z, since 7>0, so |7^{z+1}| = 7^{z+1}

So 7^{z+1} = 4

Then take log: (z+1) ln7 = ln4

So z+1 = ln4 / ln7 = log7(4)

So z = log7(4) - 1 = log7(4) - log7(7) = log7(4/7)

Which is negative, since 4/7<1.

Now look at options:

A. -2

B. -1

C. 1

D. Does not exist

Is log7(4/7) equal to -1? 7^{-1} = 1/7 ≈0.142, 4/7≈0.571, not equal.

Equal to -2? 7^{-2}=1/49≈0.0204, not 0.571.

So not integer.

But perhaps they want exact value, but options are integers or "does not exist".

Since 7^{z+1} = 4, and 7^0=1, 7^1=7, so for z+1 between 0 and 1, 7^{z+1} between 1 and 7, and 4 is in between, so there is a real solution, so D is not correct.

But the value is not among A,B,C.

Unless I misread the absolute value.

|7^{z+1}| = 4, and since 7^{z+1} >0, yes, so 7^{z+1} = 4.

Perhaps z is complex, but unlikely for this level.

Another interpretation: perhaps |7|^{z+1} = 4, but |7| =7, same thing.

Or perhaps it's 7^{|z+1|} = 4, but the absolute value is on the whole thing.

In the user's text: "|7^{z+1}| = 4", so absolute value of 7 to the power (z+1).

Yes.

Perhaps for real z, it exists, but not integer, so maybe D, but it does exist.

Unless they mean integer z, but not specified.

Let's calculate numerical value.

7^{z+1} = 4

z+1 = log7(4) = ln4/ln7 ≈ 1.3863/1.9459 ≈ 0.712

So z ≈ 0.712 -1 = -0.288, not integer.

So not A,B,C.

Perhaps the equation is |7| * |z+1| = 4, but that would be 7| z+1| =4, so |z+1| =4/7, so z+1 = ±4/7, z = -1 ±4/7, so -3/7 or -11/7, not in options.

Or if it's 7^{|z+1|} =4, then |z+1| = log7(4) ≈0.712, so z+1 = ±0.712, z = -1±0.712, so -0.288 or -1.712, not in options.

So perhaps D, "does not exist", but it does exist for real z.

Unless they mean integer z, then no integer solution, so D.

In many tests, if no integer solution, they say does not exist.

Options include "Does not exist", so perhaps that's it.

For example, if z integer, 7^{z+1} is 7^k for integer k, which is 1,7,49, etc or 1/7,1/49, etc, never 4, so no integer solution, so D.

Probably that's the intention.

So for Problem 4, answer is D.

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Back to Problem 3.

Perhaps in Problem 3, they intend for us to use the equation to find 2x - y.

From 2x + y = 2, and we have 4^{2x - y}

Let me set a = 2x, b = y, a + b = 2, want 4^{a - b}

a - b = a - (2 - a) = 2a -2, as before.

Notice that (a + b) = 2, and a - b = d, then a = (2+d)/2, b = (2-d)/2, but no constraint on d.

Perhaps the expression is \frac{16^x}{4^y} = (4^2)^x / 4^y = 4^{2x} / 4^y = 4^{2x - y}

And from 2x + y = 2, if I square or something.

Another idea: perhaps "2x + y = 2" is to be used with the expression, but maybe they want the minimum or something, but not specified.

Perhaps in the context, x and y are related, but only one equation.

Let's look at the options; 16 is there, and when y=0, x=1, it works, and perhaps that's what they assume.

Maybe the problem is "if 2x + y = 2, and x=1, find" but not said.

Perhaps it's a system, but only one equation.

Let's try to see if 2x - y can be found.

From 2x + y = 2, and if I had another equation, but I don't.

Unless the expression is constant, but it's not.

Perhaps for the value to be defined, but it is.

Another thought: perhaps "2x + y = 2" is for the exponents in a different base, but unlikely.

Let's calculate the product or sum.

Notice that 16^x / 4^y = (2^4)^x / (2^2)^y = 2^{4x - 2y} = 2^{2(2x - y)}

From 2x + y = 2, let me denote s = 2x + y = 2

d = 2x - y

Then s + d = 4x, s - d = 2y

But I have s=2, d unknown.

2^{2d} = 4^d

But d is free.

Perhaps the problem is to express in terms of s, but s is given.

I recall that in some problems, they have 2x + y = c, and ask for 4^x * 2^y or something.

For example, if it were 4^x * 2^y, then (2^2)^x * 2^y = 2^{2x} * 2^y = 2^{2x + y} = 2^2 = 4, which is option A.

Oh! Perhaps it's multiplication, not division.

In the user's text: "\frac{16^{x}}{4^{y}}" , which is division.

But maybe it's a typo, and it's multiplication.

Because if it were 16^x * 4^y, then (4^2)^x * 4^y = 4^{2x} * 4^y = 4^{2x + y} = 4^2 = 16, since 2x+y=2.

And 16 is option C.

If it were 4^x * 2^y, then 2^{2x} * 2^y = 2^{2x+y} = 2^2 = 4, option A.

But in the text, it's \frac{16^{x}}{4^{y}}, division.

However, in many fonts, / and * might be confused, but unlikely.

Perhaps in the original image, it's multiplication.

Given that, and since 16 is an option, and it makes sense, probably it's a typo, and it's multiplication.

Because otherwise, no constant value.

So for Problem 3, likely answer is 16, assuming multiplication or something.

Perhaps \frac{16^x}{4^y} = 4^{2x - y}, and from 2x + y = 2, if I add 2x - y and 2x + y, I get 4x, etc.

Let's assume that they want 4^{2x + y} = 4^2 = 16, but that's not the expression.

Another possibility: perhaps "16^x / 4^y" and they mean (16/4)^{x/y} or something, but that doesn't make sense.

I think the most reasonable is that it's a typo, and it's 16^x * 4^y or 4^{2x + y}.

Since 2x + y = 2, and if the expression was 4^{2x + y} = 4^2 = 16, or 2^{2x + y} = 4, etc.

In the options, 16 is there, and for Problem 1, if we assume it's 8!, then C.

For Problem 2, let's assume that the correct choice is not listed, but perhaps D is intended for some reason.

Let's try Problem 5.

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Problem 5: The growth of a certain bacteria in a lab is modeled by the function N_t = N_0 * e^{rt}, where N_0 represents the initial count of bacteria and N_t represents the count of bacteria after t days. After how many days will the bacteria in the lab increase by 300%?

Increase by 300% means the new amount is initial plus 300% of initial, so 4 times initial.

Because 100% + 300% = 400% = 4 times.

So N_t = 4 N_0

So 4 N_0 = N_0 * e^{rt}

Divide both sides by N_0 (assuming N_0 ≠0): 4 = e^{rt}

Take natural log: ln4 = rt

So t = ln4 / r

But r is not given, and options are numbers: A.1 B.2 C.3 D.4

So probably r is given or assumed.

In the function, r is a constant, but not specified.

Perhaps "increase by 300%" means the increase is 300% of initial, so N_t - N_0 = 3 N_0, so N_t = 4 N_0, same as above.

But t = ln4 / r, which depends on r.

Unless r is 1 or something, but not said.

Perhaps in the context, r is known, but not.

Another interpretation: "increase by 300%" might mean the rate is 300%, but that doesn't make sense.

Or perhaps r is the growth rate, and they want when it increases by 300%, but still.

Perhaps for exponential growth, the time to increase by a factor is constant, but here the factor is 4, so t = ln4 / r, and if r is given, but not.

Look at the options; perhaps they assume r= ln4 / t, but circular.

Perhaps "after how many days" and r is such that it's nice number.

For example, if r = ln4 / 2, then t=2, but why.

Perhaps in the model, r is 1, but not specified.

Another thought: perhaps "increase by 300%" means the population becomes 300% of initial, i.e., 3 times, not 4 times.

Let's check the language.

"Increase by 300%" typically means the increase is 300% of original, so new = original + 3*original = 4*original.

But sometimes people say "increases to 300%", which would be 3 times.

In common usage, "increase by 300%" means multiply by 4.

For example, if you have $100, increase by 300% means you gain $300, so have $400.

Yes.

But in some contexts, it might be ambiguous.

Perhaps for bacteria, they mean the growth rate.

Let's see the options; if N_t = 3 N_0, then 3 = e^{rt}, t = ln3 / r

Still depends on r.

Unless r is given in the problem, but it's not.

Perhaps in the function, r is implied, or perhaps it's a standard model.

Another idea: perhaps "increase by 300%" means the relative increase is 300%, but same thing.

Perhaps they mean that the growth rate r is 300% per day, but that would be r=3, then t = ln4 / 3 ≈ 1.386/3≈0.462, not in options.

If r= ln4 / 2, t=2, etc.

Perhaps for doubling time, but here it's quadrupling.

Let's calculate if r= ln2, then doubling time is 1 day, then quadrupling is 2 days, so t=2.

And 2 is option B.

Perhaps they assume that the growth rate is such that it doubles in 1 day, but not stated.

In many problems, they assume r= ln2 for doubling in 1 day, but here it's not specified.

Perhaps from the context, but no.

Another thought: perhaps "increase by 300%" means the population increases by a factor of 3, so N_t = 3 N_0.

Then 3 = e^{rt}, t = ln3 / r

If r = ln3 / 2, t=2, etc.

Still.

Perhaps the 300% is the rate, but the function has r, so r is given as 300% = 3, then t = ln4 / 3 ≈0.462, not in options.

If "increase by 300%" means the continuous growth rate is 300%, so r=3, then for increase by 300%, but increase by 300% means N_t = 4 N_0, so 4 = e^{3t}, so 3t = ln4, t = ln4 / 3 ≈ 1.386/3≈0.462, not integer.

If they mean that the population becomes 300% of initial, so N_t = 3 N_0, then 3 = e^{3t}, 3t = ln3, t = ln3 / 3 ≈ 1.0986/3≈0.366, not in options.

Perhaps "after how many days" and they want the time for the increase to be 300% of initial, but same.

Let's look at the answer choices; perhaps for t=2, if r= ln2, then N_t = N_0 e^{2 ln2} = N_0 (e^{ln2})^2 = N_0 * 4, so increase by 300%, yes.

And if they assume that the growth rate is such that it doubles in 1 day, then r = ln2, and for quadruple, t=2 days.

And 2 is option B.

Probably that's the intention.

So for Problem 5, answer is B.2

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Now Problem 6.

Problem 6: In a geotechnical survey, the soil pressure P at a depth d meters from the surface of earth is calculated by the equation P = 3 + k(d - 1), where k is a constant. If the pressure at depth 2 meters is 7, what is the pressure at depth of 4 meters?

Given P = 3 + k(d - 1)

At d=2, P=7

So 7 = 3 + k(2 - 1) = 3 + k*1 = 3 + k

So k = 7 - 3 = 4

Now at d=4, P = 3 + k(4 - 1) = 3 + 4*3 = 3 + 12 = 15

So answer is 15, option C.

Good.

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Now back to earlier problems.

For Problem 1, likely it's a typo, and it's "which is equal to 8!" , so C.8!

For Problem 2, let's assume that the correct choice is D, but earlier calculation shows not.

Perhaps in Problem 2, the expression is \left( \frac{x+1}{x-1} \right)^2, and option D is (1+x)^2, but not equal.

Another idea: perhaps "equivalent" means has the same value when x=0 or something, but not.

Let's try to see if any option equals the expression for all x.

Set \left( \frac{x+1}{x-1} \right)^2 = c * something.

Notice that \frac{x+1}{x-1} = \frac{1 + 1/x}{1 - 1/x} for x≠0, but not helpful.

Perhaps for large x, it approaches 1, while (1+x)^2 -> infinity, so not.

Let's calculate the difference.

Perhaps option B is \left( \frac{1 - x}{1 + x} \right)^2, but in user's text, it's written as B. \left[\frac{(1 - x^{2})}{(1 + x)}\right]^{2} , which is the same as A.

In the image, perhaps B is different.

Maybe B is \left( \frac{x-1}{x+1} \right)^2

At x=2, (1/3)^2=1/9, original 9, not.

Or \left( \frac{x+1}{1-x} \right)^2, which is the same as our expression, as we had.

And if that's not listed, perhaps it's implied.

But in the options, perhaps D is intended for a different reason.

Another thought: perhaps "equivalent" means can be written as that after simplification, and for example, \left( \frac{x+1}{x-1} \right)^2 = \frac{(x+1)^2}{(x-1)^2} ,