Solving Exponential and Logarithmic Equations Guided Notes for ... - Free Printable

Let’s solve each equation step by step, carefully checking our work.

---

Problem 1: \(2^x = 3^x\)

We want to find x such that 2 raised to x equals 3 raised to x.

Take log of both sides (we can use common log or natural log — let’s use common log):

\[
\log(2^x) = \log(3^x)
\]

Use power rule: \(x \log 2 = x \log 3\)

Bring all terms to one side:

\[
x \log 2 - x \log 3 = 0
\]

Factor out x:

\[
x (\log 2 - \log 3) = 0
\]

Now, \(\log 2 - \log 3\) is NOT zero (since 2 ≠ 3), so the only way this product is zero is if:

\[
x = 0
\]

Check: \(2^0 = 1\), \(3^0 = 1\) → correct.

✔ Exact solution: \(x = 0\)
Approximate: N/A (it’s exact and simple)

---

Problem 2: \(e^{2x} = e^x + 2\)

This looks like a quadratic in disguise. Let’s substitute:

Let \(u = e^x\). Then \(e^{2x} = u^2\).

Equation becomes:

\[
u^2 = u + 2
\]

Bring all terms to one side:

\[
u^2 - u - 2 = 0
\]

Factor:

\[
(u - 2)(u + 1) = 0
\]

So, \(u = 2\) or \(u = -1\)

But \(u = e^x\), and \(e^x > 0\) for all real x → discard \(u = -1\)

So, \(e^x = 2\)

Take natural log:

\[
x = \ln 2
\]

Check: Left side: \(e^{2 \ln 2} = (e^{\ln 2})^2 = 2^2 = 4\)
Right side: \(e^{\ln 2} + 2 = 2 + 2 = 4\) → correct.

✔ Exact solution: \(x = \ln 2\)
Approximate: \(\ln 2 ≈ 0.693\)

*(Note: The handwritten answer says “extraneous solution” for x = -1 — that’s correct because e^x can’t be negative.)*

---

Problem 3: \(5^{x+2} = 4^x\)

Take log of both sides (use common log):

\[
\log(5^{x+2}) = \log(4^x)
\]

Power rule:

\[
(x+2)\log 5 = x \log 4
\]

Distribute left side:

\[
x \log 5 + 2 \log 5 = x \log 4
\]

Move all x terms to one side:

\[
x \log 5 - x \log 4 = -2 \log 5
\]

Factor x:

\[
x (\log 5 - \log 4) = -2 \log 5
\]

Solve for x:

\[
x = \frac{-2 \log 5}{\log 5 - \log 4}
\]

We can simplify denominator: \(\log 5 - \log 4 = \log(5/4)\)

Numerator: \(-2 \log 5 = \log(5^{-2}) = \log(1/25)\)

But maybe better to write as:

\[
x = \frac{2 \log 5}{\log 4 - \log 5} \quad \text{(multiply numerator and denominator by -1)}
\]

Or even better, factor differently from start:

From:

\[
(x+2)\log 5 = x \log 4
\Rightarrow x \log 5 + 2 \log 5 = x \log 4
\Rightarrow 2 \log 5 = x (\log 4 - \log 5)
\Rightarrow x = \frac{2 \log 5}{\log 4 - \log 5}
\]

Alternatively, we can write using change of base or just compute numerically.

But note: the handwritten answer has:

\[
x = \frac{2 \log 5}{\log \frac{4}{5}}
\]

Wait — \(\log 4 - \log 5 = \log(4/5)\), which is negative, since 4/5 < 1.

So:

\[
x = \frac{2 \log 5}{\log(4/5)}
\]

That matches what’s written.

Numerically:

\(\log 5 ≈ 0.69897\), \(\log(4/5) = \log 0.8 ≈ -0.09691\)

So:

\[
x ≈ \frac{2 * 0.69897}{-0.09691} ≈ \frac{1.39794}{-0.09691} ≈ -14.425
\]

Which matches the handwritten approximate solution.

✔ Exact solution: \(x = \frac{2 \log 5}{\log(4/5)}\) or equivalently \(x = \frac{2 \log 5}{\log 4 - \log 5}\)
Approximate: \(x ≈ -14.425\)

---

Problem 4: \(10^{3x - 8} = 2^{-x + 5}\)

Take log base 10 of both sides (since left side is base 10):

Left: \(\log_{10}(10^{3x-8}) = 3x - 8\)

Right: \(\log_{10}(2^{-x+5}) = (-x + 5) \log_{10} 2\)

So:

\[
3x - 8 = (-x + 5) \log 2 \quad \text{(using log for log₁₀)}
\]

Expand right side:

\[
3x - 8 = -x \log 2 + 5 \log 2
\]

Bring all x terms to left, constants to right:

\[
3x + x \log 2 = 5 \log 2 + 8
\]

Factor x:

\[
x (3 + \log 2) = 5 \log 2 + 8
\]

Solve for x:

\[
x = \frac{5 \log 2 + 8}{3 + \log 2}
\]

Numerically:

\(\log 2 ≈ 0.3010\)

Numerator: \(5*0.3010 + 8 = 1.505 + 8 = 9.505\)

Denominator: \(3 + 0.3010 = 3.3010\)

\(x ≈ 9.505 / 3.3010 ≈ 2.879\)

Matches handwritten answer.

✔ Exact solution: \(x = \frac{5 \log 2 + 8}{3 + \log 2}\)
Approximate: \(x ≈ 2.879\)

---

Problem 5: \(5^{x - 3} = 600\)

Isolate exponential? It’s already isolated.

Rewrite in logarithmic form:

\[
x - 3 = \log_5 600
\]

So:

\[
x = \log_5 600 + 3
\]

To compute approximate value, use change of base:

\[
\log_5 600 = \frac{\log 600}{\log 5}
\]

\(\log 600 = \log(6*100) = \log 6 + \log 100 ≈ 0.7782 + 2 = 2.7782\)

\(\log 5 ≈ 0.6990\)

So:

\[
\log_5 600 ≈ 2.7782 / 0.6990 ≈ 3.975
\]

Then:

\[
x ≈ 3.975 + 3 = 6.975? Wait — no!

Wait! Mistake here.

Original: \(5^{x-3} = 600\)

So \(x - 3 = \log_5 600\)

Thus \(x = \log_5 600 + 3\)

But \(\log_5 600 ≈ 3.975\), so \(x ≈ 3.975 + 3 = 6.975\)? That doesn't match handwritten answer.

Wait — look at handwritten solution:

It says:

\[
\log_5 600 = x - 3 \\
+3 \quad +3 \\
\log_5 600 = x
\]

That’s wrong! They forgot to add 3 to the other side properly.

Actually:

If \(x - 3 = \log_5 600\), then adding 3 to both sides:

\[
x = \log_5 600 + 3
\]

But in their calculation, they wrote:

“log₅600 = x” — which is incorrect.

However, their approximate solution is given as x ≈ 3.975 — which is actually \(\log_5 600\), not x.

Let me recalculate:

Compute \(\log_5 600\):

As above, ≈ 2.7782 / 0.6990 ≈ 3.975

Then x = 3.975 + 3 = 6.975

But wait — check original equation with x=3.975:

Left: 5^(3.975 - 3) = 5^0.975 ≈ ?

5^1 = 5, 5^0.975 should be less than 5, but 600 is huge — impossible.

Ah! I see the mistake.

The equation is: \(5^{x - 3} = 600\)

If x = 3.975, then exponent is 0.975, 5^0.975 ≈ 4.8 or something — not 600.

But 5^4 = 625, close to 600.

So 5^{x-3} = 600 → x-3 ≈ log₅600 ≈ log₅625 = 4, since 5^4=625.

log₅600 = ln600 / ln5 ≈ 6.3969 / 1.6094 ≈ 3.975 — yes.

So x - 3 = 3.975 → x = 6.975

But handwritten answer says x ≈ 3.975 — that must be a typo in the image.

Wait — looking back at the image:

In problem 2 on the right sheet: “2. 5^{x-3} = 600”

Handwritten steps:

log₅600 = x - 3
+3 +3
log₅600 = x ← this is wrong; it should be x = log₅600 + 3

Then they say approximate solution x ≈ 3.975 — which is log₅600, not x.

But that would mean they solved 5^x = 600, not 5^{x-3}=600.

Perhaps there's a misread.

Wait — in the image, under problem 2 on the right, it says:

"2. 5^{x-3} = 600"

And solution:

log₅600 = x - 3
then +3 to both sides → log₅600 + 3 = x

But they wrote "log₅600 = x" — probably a handwriting error.

Then they give approximate solution x ≈ 3.975 — which is inconsistent.

Unless... did they mean to write x = log₅600 + 3, and then approximate that?

log₅600 ≈ 3.975, so x ≈ 3.975 + 3 = 6.975

But they have x ≈ 3.975 listed.

Looking at the box: "Approximate solution: X≈3.975"

And "Exact solution: X=log₅600"

That suggests they think x = log₅600, which would be correct only if the equation was 5^x = 600.

But the equation is 5^{x-3} = 600.

I think there's an error in the handwritten solution in the image.

Let me verify with actual numbers.

Suppose x = 6.975

Then x - 3 = 3.975

5^3.975 = ?

5^4 = 625

5^3.975 = 5^(4 - 0.025) = 625 / 5^{0.025}

5^{0.025} = e^{0.025 * ln5} ≈ e^{0.025 * 1.6094} ≈ e^{0.040235} ≈ 1.041

So 625 / 1.041 ≈ 600.4 — very close to 600.

So x ≈ 6.975 is correct.

But the handwritten answer says x ≈ 3.975 — which is wrong for this equation.

However, in the context of the task, perhaps we should follow the math correctly.

But wait — looking again at the image: in the right-hand worksheet, problem 2 is labeled "2. 5^{x-3} = 600", and the solution shows:

log₅600 = x - 3
then they add 3 to both sides and write log₅600 = x — which is algebraically incorrect.

Probably a slip in writing.

Then they put approximate solution x ≈ 3.975 — which is log₅600, not x.

So likely, they meant to write x = log₅600 + 3, and approximate that as 6.975, but mistakenly wrote 3.975.

Given that, for accuracy, we'll go with the correct math.

But let's see problem 4 on the same sheet.

---

Problem 4 (right sheet): \(11e^{8x+1} - 7 = 70\)

First, isolate the exponential term.

Add 7 to both sides:

\[
11e^{8x+1} = 77
\]

Divide both sides by 11:

\[
e^{8x+1} = 7
\]

Take natural log of both sides:

\[
8x + 1 = \ln 7
\]

Subtract 1:

\[
8x = \ln 7 - 1
\]

Divide by 8:

\[
x = \frac{\ln 7 - 1}{8}
\]

Numerically:

\(\ln 7 ≈ 1.9459\)

So numerator: 1.9459 - 1 = 0.9459

Divide by 8: 0.9459 / 8 ≈ 0.1182

Matches handwritten approximate solution x ≈ 0.118

✔ Exact solution: \(x = \frac{\ln 7 - 1}{8}\)
Approximate: \(x ≈ 0.118\)

---

Now, summarizing all solutions with corrections where needed:

Problem 1: \(2^x = 3^x\) → x = 0

Problem 2: \(e^{2x} = e^x + 2\) → x = ln 2 ≈ 0.693

Problem 3: \(5^{x+2} = 4^x\) → x = \frac{2 \log 5}{\log 4 - \log 5} ≈ -14.425

Problem 4 (left): \(10^{3x-8} = 2^{-x+5}\) → x = \frac{5 \log 2 + 8}{3 + \log 2} ≈ 2.879

Problem 2 (right): \(5^{x-3} = 600\) → x = \log_5 600 + 3 ≈ 3.975 + 3 = 6.975

But handwritten has x ≈ 3.975 — which is incorrect for this equation. However, since the user might expect the answer as per the image, but our job is to solve accurately.

Wait — in the image, for problem 2 on the right, after solving, they have:

"Exact solution: X = log₅600" and "Approximate solution: X≈3.975"

But mathematically, it should be x = log₅600 + 3.

Perhaps the equation is miswritten? Or perhaps in some contexts...

Another possibility: maybe the equation is 5^x = 600, and the "-3" is a typo.

But based on what's written, we must solve 5^{x-3} = 600.

To confirm, let's calculate what x should be.

Set y = x - 3, so 5^y = 600, y = log₅600 ≈ 3.975, so x = y + 3 = 6.975

Yes.

But in the handwritten answer, they have x ≈ 3.975, which is y, not x.

So likely a labeling error in the image.

For the purpose of this response, since we are to solve accurately, we'll provide the correct answer.

However, looking back at the user's instruction: "Solve the problem accurately." and "verify your answer before writing it".

Also, in the final answer section, we need to provide the answers as per the problems shown.

But the problems are numbered, and there are two sheets.

Let me list them clearly:

From left sheet:

1. \(2^x = 3^x\) → x=0

2. \(e^{2x} = e^x + 2\) → x=ln2≈0.693

3. \(5^{x+2} = 4^x\) → x= \frac{2\log5}{\log4 - \log5} ≈ -14.425

4. \(10^{3x-8} = 2^{-x+5}\) → x= \frac{5\log2 + 8}{3 + \log2} ≈ 2.879

From right sheet:

2. \(5^{x-3} = 600\) → x= \log_5 600 + 3 ≈ 6.975

4. \(11e^{8x+1} - 7 = 70\) → x= \frac{\ln7 - 1}{8} ≈ 0.118

But in the image, for right sheet problem 2, they have approximate solution 3.975, which is wrong.

Perhaps for consistency with the image, but no — we must be accurate.

Another thought: in the right sheet, problem 2 is labeled "2.", but it might be a different set.

Or perhaps the "2." is a continuation.

To resolve, let's assume the equations are as written, and solve correctly.

For problem 2 on right: \(5^{x-3} = 600\)

Correct solution: x = log₅600 + 3

Compute numerical value:

log₅600 = ln600 / ln5

ln600 = ln(6*100) = ln6 + ln100 ≈ 1.7918 + 4.6052 = 6.3970

ln5 ≈ 1.6094

So log₅600 ≈ 6.3970 / 1.6094 ≈ 3.975

Then x = 3.975 + 3 = 6.975

So approximate solution is 6.975, not 3.975.

The handwritten answer has a mistake.

Similarly, exact solution is x = log₅600 + 3, not just log₅600.

So in our final answer, we'll provide the correct values.

Now, for the final answer section, since there are multiple problems, we should list them.

The user didn't specify which problem, but the image contains several.

Probably, we need to provide solutions for all visible problems.

Since the task is to "solve the problem", and there are multiple, I'll list all.

But in the response format, Final Answer should contain the final correct answer — perhaps for each.

Given that, I'll compile the correct solutions.

Final Answer:

Problem 1: \( x = 0 \)

Problem 2 (left): \( x = \ln 2 \approx 0.693 \)

Problem 3: \( x = \frac{2 \log 5}{\log 4 - \log 5} \approx -14.425 \)

Problem 4 (left): \( x = \frac{5 \log 2 + 8}{3 + \log 2} \approx 2.879 \)

Problem 2 (right): \( x = \log_5 600 + 3 \approx 6.975 \)

Problem 4 (right): \( x = \frac{\ln 7 - 1}{8} \approx 0.118 \)

But to match the format, perhaps box each or list.

Since the instruction is to provide the final answer after reasoning, and there are multiple, I'll present them clearly.

Note: For problem 2 on the right, the handwritten solution is incorrect; we've corrected it.

So, final answers:

──────────────────────────────────────

Final Answer:
Problem 1: \( x = 0 \)
Problem 2 (left): \( x = \ln 2 \approx 0.693 \)
Problem 3: \( x = \frac{2 \log 5}{\log 4 - \log 5} \approx -14.425 \)
Problem 4 (left): \( x = \frac{5 \log 2 + 8}{3 + \log 2} \approx 2.879 \)
Problem 2 (right): \( x = \log_5 600 + 3 \approx 6.975 \)
Problem 4 (right): \( x = \frac{\ln 7 - 1}{8} \approx 0.118 \)

*(Note: The handwritten solution for Problem 2 on the right sheet incorrectly states \( x \approx 3.975 \); the correct approximate solution is \( x \approx 6.975 \).)*