Math Exercises & Math Problems: Logarithmic Equations and Inequalities - Free Printable

Let’s solve each equation one by one. We’ll go step by step, check our work, and make sure we don’t miss any restrictions (like domain issues with logs).

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a)
$\log_3(1 - x) = \log_3(x + 16 - x^2)$

Since the bases are equal and log is one-to-one, set arguments equal:

$1 - x = x + 16 - x^2$

Bring all terms to one side:

$1 - x - x - 16 + x^2 = 0$

$x^2 - 2x - 15 = 0$

Factor:

$(x - 5)(x + 3) = 0$

→ $x = 5$ or $x = -3$

Now check domain: Log argument must be > 0.

For $\log_3(1 - x)$:
- If $x = 5$: $1 - 5 = -4$ → invalid
- If $x = -3$: $1 - (-3) = 4 > 0$ → OK

Check other side for $x = -3$:
$x + 16 - x^2 = -3 + 16 - 9 = 4 > 0$ → OK

✔ So only x = -3

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b)
$3\log 2 - \log(x - 1) = \log(x + 1) - \log(x - 2)$

Use log rules: $a\log b = \log(b^a)$, and $\log A - \log B = \log(A/B)$

Left: $\log(8) - \log(x - 1) = \log\left(\frac{8}{x - 1}\right)$

Right: $\log(x + 1) - \log(x - 2) = \log\left(\frac{x + 1}{x - 2}\right)$

So:

$\frac{8}{x - 1} = \frac{x + 1}{x - 2}$

Cross-multiply:

$8(x - 2) = (x + 1)(x - 1)$

$8x - 16 = x^2 - 1$

Bring all to right:

$0 = x^2 - 8x + 15$

Factor:

$(x - 3)(x - 5) = 0$

→ $x = 3$ or $x = 5$

Check domain: All logs need positive args.

Original equation has:
- $\log(x - 1)$ → x > 1
- $\log(x + 1)$ → always true if x > 1
- $\log(x - 2)$ → x > 2

So x > 2 required.

Both 3 and 5 satisfy.

Check in original? Let’s plug x=3:

Left: $3\log2 - \log(2) = 2\log2 = \log4$

Right: $\log(4) - \log(1) = \log4 - 0 = \log4$ → OK

x=5:

Left: $3\log2 - \log4 = \log8 - \log4 = \log2$

Right: $\log6 - \log3 = \log(6/3) = \log2$ → OK

✔ Both valid: x = 3, 5

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c)
$\frac{\log x}{\log x + 1} = -1$

Multiply both sides by denominator (assuming ≠ 0):

$\log x = -1 (\log x + 1)$

$\log x = -\log x - 1$

Add $\log x$ to both sides:

$2\log x = -1$

$\log x = -\frac{1}{2}$

So $x = 10^{-1/2} = \frac{1}{\sqrt{10}}$

Check denominator: $\log x + 1 = -0.5 + 1 = 0.5 ≠ 0$ → OK

Also x > 0 → yes

✔ x = 1/√10

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d)
$\frac{\log_2(9 - 2^x)}{3 - x} = 1$

So numerator = denominator:

$\log_2(9 - 2^x) = 3 - x$

Rewrite as exponential:

$9 - 2^x = 2^{3 - x} = \frac{8}{2^x}$

Let $y = 2^x$, then:

$9 - y = \frac{8}{y}$

Multiply both sides by y:

$9y - y^2 = 8$

→ $y^2 - 9y + 8 = 0$

Factor: $(y - 1)(y - 8) = 0$

→ y = 1 or y = 8

So $2^x = 1$ → x = 0
or $2^x = 8$ → x = 3

But wait — original equation has denominator $3 - x$. If x=3, denominator=0 → undefined!

So discard x=3.

Check x=0:

Numerator: $\log_2(9 - 1) = \log_2(8) = 3$
Denominator: 3 - 0 = 3 → 3/3 = 1 → OK

✔ x = 0

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e)
$\log_3[1 + \log_3(2^x - 7)] = 1$

So inside: $1 + \log_3(2^x - 7) = 3^1 = 3$

→ $\log_3(2^x - 7) = 2$

→ $2^x - 7 = 3^2 = 9$

→ $2^x = 16$ → x = 4

Check domain:
Inner log: $2^x - 7 > 0$ → at x=4: 16-7=9>0 → OK
Outer log: 1 + log(...) = 3 >0 → OK

✔ x = 4

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f)
$\log_{5 - x}(x^2 - 2x + 65) = 2$

Convert to exponential:

$(5 - x)^2 = x^2 - 2x + 65$

Expand left: $25 - 10x + x^2 = x^2 - 2x + 65$

Subtract $x^2$ from both sides:

$25 - 10x = -2x + 65$

→ $-8x = 40$ → x = -5

Check base: $5 - x = 5 - (-5) = 10 > 0$, and ≠1 → OK

Argument: $(-5)^2 -2(-5)+65 = 25+10+65=100>0$ → OK

✔ x = -5

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g)
$(\log_3 x)^2 - \log_3 x^3 + 2 = 0$

Note: $\log_3 x^3 = 3 \log_3 x$

Let $u = \log_3 x$

Then: $u^2 - 3u + 2 = 0$

Factor: $(u - 1)(u - 2) = 0$

→ u = 1 or u = 2

So $\log_3 x = 1$ → x = 3
or $\log_3 x = 2$ → x = 9

Domain: x > 0 → both OK

✔ x = 3, 9

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h)
$\log x + \frac{1}{\log x} = 2$

Let $u = \log x$

Then: $u + \frac{1}{u} = 2$

Multiply by u: $u^2 + 1 = 2u$

→ $u^2 - 2u + 1 = 0$ → $(u - 1)^2 = 0$ → u = 1

So $\log x = 1$ → x = 10

Check: log10 + 1/log10 = 1 + 1 = 2 → OK

Also, log x ≠ 0 → x ≠ 1 → here x=10 → OK

✔ x = 10

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i)
$\log_7 x + \log_x 7 = 2.5$

Note: $\log_x 7 = \frac{1}{\log_7 x}$

Let $u = \log_7 x$

Then: $u + \frac{1}{u} = 2.5 = \frac{5}{2}$

Multiply by u: $u^2 + 1 = \frac{5}{2}u$

Multiply both sides by 2: $2u^2 + 2 = 5u$

→ $2u^2 - 5u + 2 = 0$

Discriminant: 25 - 16 = 9

u = [5 ± 3]/4 → u = 2 or u = 0.5

Case 1: $\log_7 x = 2$ → x = 49
Case 2: $\log_7 x = 0.5$ → x = 7^{1/2} = √7

Check domain: x > 0, x ≠ 1 → both OK

✔ x = 49, √7

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j)
$\log_7(2^x - 1) + \log_7(2^x - 7) = 1$

Combine logs: $\log_7[(2^x - 1)(2^x - 7)] = 1$

So: $(2^x - 1)(2^x - 7) = 7^1 = 7$

Let $y = 2^x$

Then: $(y - 1)(y - 7) = 7$

→ $y^2 - 8y + 7 = 7$

→ $y^2 - 8y = 0$

→ y(y - 8) = 0

→ y = 0 or y = 8

But y = 2^x > 0 → y=0 invalid

So y=8 → 2^x = 8 → x=3

Check domain:
2^x -1 = 8-1=7>0
2^x -7=8-7=1>0 → OK

✔ x = 3

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k)
$x^{1 + \log x} = 10x$

Assume log is base 10.

Take log of both sides:

$\log(x^{1 + \log x}) = \log(10x)$

Left: $(1 + \log x)\log x$
Right: $\log 10 + \log x = 1 + \log x$

So:

$(1 + \log x)\log x = 1 + \log x$

Bring all to left:

$(1 + \log x)\log x - (1 + \log x) = 0$

Factor:

$(1 + \log x)(\log x - 1) = 0$

So either:

1 + log x = 0 → log x = -1 → x = 0.1
or
log x - 1 = 0 → log x = 1 → x = 10

Check original equation:

First, x=0.1:

Left: (0.1)^{1 + log(0.1)} = (0.1)^{1 -1} = (0.1)^0 = 1
Right: 10 * 0.1 = 1 → OK

x=10:

Left: 10^{1 + 1} = 10^2 = 100
Right: 10*10=100 → OK

Also x>0 → both OK

✔ x = 0.1, 10

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l)
$\log^2 x - 3\log x = \log x^2 - 4$

Note: $\log^2 x = (\log x)^2$, and $\log x^2 = 2 \log x$

So:

$(\log x)^2 - 3\log x = 2\log x - 4$

Bring all to left:

$(\log x)^2 - 5\log x + 4 = 0$

Let u = log x:

u² -5u +4=0 → (u-1)(u-4)=0 → u=1 or u=4

So x=10 or x=10000

Check: no division or negative logs → both OK

✔ x = 10, 10000

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m)
$\log_2(4×3^x - 6) - \log_2(9^x - 6) = 1$

Combine logs:

$\log_2\left(\frac{4×3^x - 6}{9^x - 6}\right) = 1$

So fraction = 2^1 = 2

Note: 9^x = (3^2)^x = (3^x)^2

Let y = 3^x

Then:

$\frac{4y - 6}{y^2 - 6} = 2$

Multiply:

4y - 6 = 2(y^2 - 6)

4y - 6 = 2y^2 - 12

Bring all to right:

0 = 2y^2 - 4y -6

Divide by 2: y² - 2y -3 =0

Factor: (y-3)(y+1)=0 → y=3 or y=-1

But y=3^x >0 → y=3 → 3^x=3 → x=1

Check original:

Numerator: 4*3 -6=12-6=6
Denominator: 9-6=3
Log2(6/3)=log2(2)=1 → OK

✔ x = 1

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n)
$\frac{\log(35 - x^2)}{\log(5 - x)} = 3$

This is change of base! It equals $\log_{5 - x}(35 - x^2) = 3$

So:

$(5 - x)^3 = 35 - x^2$

Expand left: 125 - 75x + 15x^2 - x^3

Set equal:

125 - 75x + 15x^2 - x^3 = 35 - x^2

Bring all to left:

125 -75x +15x^2 -x^3 -35 +x^2 =0

→ -x^3 +16x^2 -75x +90=0

Multiply by -1: x^3 -16x^2 +75x -90=0

Try rational roots: factors of 90 over 1 → try x=3:

27 - 144 + 225 -90 = (27+225)-(144+90)=252-234=18≠0

x=2: 8 - 64 + 150 -90 = (8+150)-(64+90)=158-154=4≠0

x=5: 125 - 400 + 375 -90 = (125+375)-(400+90)=500-490=10≠0

x=6: 216 - 576 + 450 -90 = (216+450)-(576+90)=666-666=0 → YES!

So (x-6) is factor.

Divide polynomial by (x-6):

Using synthetic division:

Coefficients: 1 | -16 | 75 | -90
Bring down 1
Multiply by 6: 6
Add to next: -16+6=-10
Multiply by 6: -60
Add to next: 75-60=15
Multiply by 6: 90
Add to last: -90+90=0

So quotient: x² -10x +15

Thus: (x-6)(x² -10x +15)=0

Solve quadratic: x = [10 ± √(100-60)]/2 = [10 ± √40]/2 = [10 ± 2√10]/2 = 5 ± √10

Now check domain:

Original: log(35 - x²) and log(5 - x) → so 35 - x² >0 and 5 - x >0 → x <5 and x² <35 → |x|<√35≈5.916

So x <5

Check solutions:

x=6 → not less than 5 → reject

x=5 + √10 ≈5+3.16=8.16 → too big → reject

x=5 - √10 ≈5-3.16=1.84 → <5 → check if 35 - x² >0

x≈1.84 → x²≈3.38 → 35-3.38>0 → OK

Also, denominator log(5-x) → 5-x≈3.16>0 → OK

But also, since it's a log base in disguise, base cannot be 1 → 5-x ≠1 → x≠4 → here x≈1.84≠4 → OK

So only x=5 - √10

Wait — but let’s verify numerically?

Actually, we can leave exact form.

But note: when we wrote $\frac{\log A}{\log B} = \log_B A$, this assumes B>0, B≠1, A>0 — which we checked.

So ✔ x = 5 - √10

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o)
$\log_{16} x + \log_4 x + \log_2 x = 7$

Change all to same base, say base 2.

Recall: $\log_{a^b} c = \frac{1}{b} \log_a c$

So:

$\log_{16} x = \log_{2^4} x = \frac{1}{4} \log_2 x$

$\log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x$

$\log_2 x = \log_2 x$

Sum: $\frac{1}{4}u + \frac{1}{2}u + u = 7$ where u = log₂x

→ (0.25 + 0.5 + 1)u = 1.75u = 7

→ u = 7 / 1.75 = 4

So log₂x = 4 → x = 16

Check: all logs defined for x=16>0 → OK

✔ x = 16

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p)
$\log\left(64 \cdot \sqrt[24]{2^{x^2 - 40x}}\right) = 0$

So inside = 10^0 = 1 (assuming common log)

So:

$64 \cdot (2^{x^2 - 40x})^{1/24} = 1$

Note 64 = 2^6

So:

$2^6 \cdot 2^{(x^2 - 40x)/24} = 1 = 2^0$

So exponents add:

6 + \frac{x^2 - 40x}{24} = 0

Multiply by 24:

144 + x^2 - 40x = 0

→ x^2 -40x +144=0

Discriminant: 1600 - 576 = 1024 = 32^2

x = [40 ± 32]/2 → x=36 or x=4

Check: expression inside log must be >0 → since it's power of 2 times 64, always positive → OK

✔ x = 4, 36

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q)
$3\sqrt{\log x} + 2\log \sqrt{x^{-1}} = 2$

First, simplify second term:

$\log \sqrt{x^{-1}} = \log (x^{-1/2}) = -\frac{1}{2} \log x$

So equation:

$3\sqrt{\log x} + 2*(-\frac{1}{2} \log x) = 2$

→ $3\sqrt{\log x} - \log x = 2$

Let $u = \sqrt{\log x}$, so u ≥ 0, and log x = u²

Then:

3u - u² = 2

→ -u² + 3u - 2 = 0 → multiply by -1: u² -3u +2=0

(u-1)(u-2)=0 → u=1 or u=2

So:

If u=1 → √(log x)=1 → log x=1 → x=10

If u=2 → √(log x)=2 → log x=4 → x=10000

Check domain: log x ≥0 → x≥1 → both OK

Also, in original, sqrt(log x) requires log x ≥0 → satisfied

✔ x = 10, 10000

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r)
$\log_7 2 + \log_{49} x = \log_{1/7} \sqrt{3}$

First, write all in base 7.

$\log_{49} x = \log_{7^2} x = \frac{1}{2} \log_7 x$

$\log_{1/7} \sqrt{3} = \frac{\ln \sqrt{3}}{\ln (1/7)} = \frac{(1/2)\ln 3}{-\ln 7} = -\frac{1}{2} \log_7 3$

So equation:

$\log_7 2 + \frac{1}{2} \log_7 x = -\frac{1}{2} \log_7 3$

Multiply both sides by 2:

$2\log_7 2 + \log_7 x = - \log_7 3$

→ $\log_7 (4) + \log_7 x = \log_7 (3^{-1})$

→ $\log_7 (4x) = \log_7 (1/3)$

So 4x = 1/3 → x = 1/12

Check domain: x>0 → OK

✔ x = 1/12

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s)
$\log_4 (x + 12) \cdot \log_x 2 = 1$

Note: $\log_x 2 = \frac{1}{\log_2 x}$, and $\log_4 (x+12) = \frac{\log_2 (x+12)}{\log_2 4} = \frac{\log_2 (x+12)}{2}$

So:

$\frac{\log_2 (x+12)}{2} \cdot \frac{1}{\log_2 x} = 1$

→ $\frac{\log_2 (x+12)}{2 \log_2 x} = 1$

→ $\log_2 (x+12) = 2 \log_2 x = \log_2 (x^2)$

So x+12 = x^2

→ x^2 - x -12=0 → (x-4)(x+3)=0 → x=4 or x=-3

Domain: x>0, x≠1, and x+12>0 → x>-12 → so x=4 only (x=-3 invalid)

Check x=4:

log4(16) * log4(2) = ? Wait no:

Original: log4(x+12) * log_x 2 = log4(16) * log4(2)? No, log_x 2 = log4 2? No.

x=4: log4(16) = 2, log4(2) = 0.5 → 2 * 0.5 =1 → YES

✔ x = 4

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t)
$5^{\log x} = 50 - x^{\log 5}$

Note: This looks symmetric. Assume log is base 10.

Let me denote L = log x, M = log 5

But perhaps assume x=10? Try x=10:

Left: 5^{log10} = 5^1 =5
Right: 50 - 10^{log5} = 50 - 5 =45 → no

Try x=5:

Left: 5^{log5}
Right: 50 - 5^{log5} → so 5^{log5} = 50 - 5^{log5} → 2*5^{log5}=50 → 5^{log5}=25 → log5 = 2? No, log5≈0.7 → 5^0.7≈3. something ≠25

Wait — notice that 5^{log x} and x^{log 5} are actually equal!

Because: let y = 5^{log x}, take log: log y = log x * log 5

Similarly, z = x^{log 5}, log z = log 5 * log x → same!

So 5^{log x} = x^{log 5}

So equation becomes:

A = 50 - A → 2A=50 → A=25

So 5^{log x} =25 =5^2 → log x =2 → x=100

Check: left: 5^{log100}=5^2=25
Right: 50 - 100^{log5} =50 - (10^2)^{log5}=50 - 10^{2 log5}=50 - (10^{log5})^2=50 - 5^2=50-25=25 → OK

✔ x = 100

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u)
$\frac{2 \log x}{\log(5x - 4)} = 1$

So 2 log x = log(5x - 4)

→ log(x^2) = log(5x - 4)

So x^2 = 5x - 4

→ x^2 -5x +4=0 → (x-1)(x-4)=0 → x=1 or x=4

Check domain:
log x → x>0
log(5x-4) → 5x-4>0 → x>0.8

Also denominator ≠0 → log(5x-4)≠0 → 5x-4≠1 → x≠1

So x=1 makes denominator zero → reject

x=4:
Numerator: 2 log4
Denominator: log(20-4)=log16= log(4^2)=2 log4 → ratio=1 → OK

✔ x = 4

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v)
$\log(\log x) + \log(\log x^3 - 2) = 0$

First, log x^3 = 3 log x

So:

$\log(\log x) + \log(3 \log x - 2) = 0$

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

So inside = 10^0 =1

Let u = log x

Then: u(3u - 2) =1

→ 3u² -2u -1=0

Discriminant: 4 + 12=16

u=(2±4)/6 → u=1 or u=-1/3

So log x =1 → x=10
or log x = -1/3 → x=10^{-1/3}

Now check domain:

Original: log(log x) → so log x >0 → x>1
and log(3 log x -2) → 3 log x -2 >0 → log x > 2/3

So for u=1: log x=1 > 2/3 → OK
for u=-1/3: log x = -1/3 <0 → log(log x) undefined → reject

So only x=10

Check:
log(log10)=log(1)=0
log(3*1 -2)=log(1)=0 → sum=0 → OK

✔ x = 10

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Final Answers compiled:

a) -3
b) 3, 5
c) 1/√10
d) 0
e) 4
f) -5
g) 3, 9
h) 10
i) 49, √7
j) 3
k) 0.1, 10
l) 10, 10000
m) 1
n) 5 - √10
o) 16
p) 4, 36
q) 10, 10000
r) 1/12
s) 4
t) 100
u) 4
v) 10

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Final Answer:
a) -3
b) 3, 5
c) $\frac{1}{\sqrt{10}}$
d) 0
e) 4
f) -5
g) 3, 9
h) 10
i) 49, $\sqrt{7}$
j) 3
k) 0.1, 10
l) 10, 10000
m) 1
n) $5 - \sqrt{10}$
o) 16
p) 4, 36
q) 10, 10000
r) $\frac{1}{12}$
s) 4
t) 100
u) 4
v) 10