Collection of logarithmic equations for solving.

Let's solve each logarithmic equation step by step. We'll go through all the problems from a) to v) and explain the solution process for each.

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a) $\log_x 2 = 2$

This means:
$$
x^2 = 2 \Rightarrow x = \sqrt{2}
$$

Answer: $x = \sqrt{2}$

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b) $\log_2 x = 10$

Rewrite in exponential form:
$$
x = 2^{10} = 1024
$$

Answer: $x = 1024$

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c) $\log x^2 = 2$

Assuming base 10 (common log):
$$
\log_{10}(x^2) = 2 \Rightarrow x^2 = 10^2 = 100 \Rightarrow x = \pm10
$$

But $\log(x^2)$ is defined for $x \ne 0$, so both values are acceptable, but we must check domain of original expression.

Wait — actually, $\log x^2 = 2$ implies $x^2 = 100$, so $x = \pm10$. But $\log x^2$ is defined as long as $x^2 > 0$, which it is for $x = \pm10$. So both valid.

However, sometimes people interpret $\log x^2$ as $2\log|x|$, so yes, both are valid.

Answer: $x = \pm10$

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d) $\log_9 x^2 = 9$

$$
x^2 = 9^9 \Rightarrow x = \pm 9^{9/2} = \pm (3^2)^{9/2} = \pm 3^9 = \pm 19683
$$

Answer: $x = \pm 19683$

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e) $\log_5 25 = 2$

Check if true:
$$
5^2 = 25 \Rightarrow \text{True}
$$

So this is an identity — it's already true.

But the problem says "solve", so maybe it's just confirming?

Wait — it's written as an equation: $\log_5 25 = 2$. This is true, so no variable to solve.

But likely it's a typo or mislabeling. Perhaps it was meant to be something like $\log_5 x = 2$? But as written, it's just a statement.

Answer: True (no variable to solve)

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f) $\log_3 21 = 7$

Check: $3^7 = 2187 \ne 21$, so false.

So this equation is not true.

But again, it's written as an equation. If it's asking to solve, then:

Wait — it's probably not an equation to solve, but rather a statement that might be false.

But since it's listed among others with variables, perhaps it's a typo.

Alternatively, maybe it's supposed to be $\log_3 x = 7$, then $x = 3^7 = 2187$

But as written: $\log_3 21 = 7$ → False.

So unless it's asking to verify, it's not solvable.

But likely f) is meant to be: $\log_3 x = 7$, then $x = 3^7 = 2187$

Assume typo.

Answer (if $\log_3 x = 7$): $x = 2187$

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g) $\log_x 128 = \frac{1}{2}$

This means:
$$
x^{1/2} = 128 \Rightarrow \sqrt{x} = 128 \Rightarrow x = 128^2 = 16384
$$

Check: $\log_{16384} 128 = \frac{1}{2}$?

Yes, because $(16384)^{1/2} = 128$

Answer: $x = 16384$

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h) $\log_{81} x = -1$

$$
x = 81^{-1} = \frac{1}{81}
$$

Answer: $x = \frac{1}{81}$

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i) $\log_4 x = 3$

$$
x = 4^3 = 64
$$

Answer: $x = 64$

---

j) $\log_x 4 = 3$

$$
x^3 = 4 \Rightarrow x = \sqrt[3]{4}
$$

Answer: $x = 4^{1/3}$

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k) $\log x - \log 5 = 2$

Assume base 10:
$$
\log\left(\frac{x}{5}\right) = 2 \Rightarrow \frac{x}{5} = 10^2 = 100 \Rightarrow x = 500
$$

Answer: $x = 500$

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l) $\log_4 4 + \log_2 2 = 1$

Evaluate:
- $\log_4 4 = 1$
- $\log_2 2 = 1$
So: $1 + 1 = 2 \ne 1$

So this is false.

But again, it's written as an equation. Is it asking to solve? There’s no variable.

Possibly a typo. Maybe it should be:

$$
\log_4 x + \log_2 2 = 1
\Rightarrow \log_4 x + 1 = 1 \Rightarrow \log_4 x = 0 \Rightarrow x = 1
$$

But as written: False

Answer: Not true (no solution)

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m) $\log(x - 5) - \log(1 - x) = \frac{1}{3}$

Use log rule:
$$
\log\left(\frac{x - 5}{1 - x}\right) = \frac{1}{3}
\Rightarrow \frac{x - 5}{1 - x} = 10^{1/3}
$$

Note: Domain requires:
- $x - 5 > 0 \Rightarrow x > 5$
- $1 - x > 0 \Rightarrow x < 1$

Contradiction! No $x$ satisfies both.

So no solution

Answer: No solution (domain contradiction)

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n) $\log_5 2 + 2\log_5 x = \log_5 18$

Use log rules:
$$
\log_5 2 + \log_5 x^2 = \log_5 18 \Rightarrow \log_5 (2x^2) = \log_5 18
\Rightarrow 2x^2 = 18 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3
$$

But $\log_5 x$ is only defined for $x > 0$, so $x = 3$

Answer: $x = 3$

---

o) $\log_4(4x + 8) = 2$

$$
4x + 8 = 4^2 = 16 \Rightarrow 4x = 8 \Rightarrow x = 2
$$

Check: $4(2)+8=16$, $\log_4 16 = 2$ → Yes

Answer: $x = 2$

---

p) $\log_3(x^2 - 8x) = 2$

$$
x^2 - 8x = 3^2 = 9 \Rightarrow x^2 - 8x - 9 = 0
\Rightarrow (x - 9)(x + 1) = 0 \Rightarrow x = 9 \text{ or } x = -1
$$

Check domain: $x^2 - 8x > 0$

For $x = 9$: $81 - 72 = 9 > 0$ → OK
For $x = -1$: $1 + 8 = 9 > 0$ → OK

Both satisfy.

Answer: $x = 9$ or $x = -1$

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q) $\log_3 81 - 0.5 = \log_3 27$

Compute:
- $\log_3 81 = \log_3(3^4) = 4$
- $\log_3 27 = \log_3(3^3) = 3$

So: $4 - 0.5 = 3.5 \ne 3$ → False

But again, no variable. Likely a verification.

Answer: False

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r) $\frac{\log x}{\log(5x - 3)} = 1$

So:
$$
\log x = \log(5x - 3) \Rightarrow x = 5x - 3 \Rightarrow -4x = -3 \Rightarrow x = \frac{3}{4}
$$

Check domain:
- $x > 0$: $3/4 > 0$ → OK
- $5x - 3 > 0$: $5*(3/4) = 15/4 = 3.75 > 3$ → $3.75 - 3 = 0.75 > 0$ → OK

So valid.

Answer: $x = \frac{3}{4}$

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s) $\frac{2 + \log x}{3 - \log x} = 5$

Let $y = \log x$, then:
$$
\frac{2 + y}{3 - y} = 5 \Rightarrow 2 + y = 5(3 - y) = 15 - 5y
\Rightarrow y + 5y = 15 - 2 \Rightarrow 6y = 13 \Rightarrow y = \frac{13}{6}
$$

So $\log x = \frac{13}{6} \Rightarrow x = 10^{13/6} = 10^{2 + 1/6} = 100 \cdot 10^{1/6}$

Or leave as $x = 10^{13/6}$

Answer: $x = 10^{13/6}$

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

Use log rule:
$$
\log\left(\frac{3x^2 + 1}{3 + x}\right) = \log(3x - 2)
\Rightarrow \frac{3x^2 + 1}{x + 3} = 3x - 2
$$

Multiply both sides:
$$
3x^2 + 1 = (3x - 2)(x + 3) = 3x(x) + 9x - 2x - 6 = 3x^2 + 7x - 6
$$

So:
$$
3x^2 + 1 = 3x^2 + 7x - 6 \Rightarrow 1 = 7x - 6 \Rightarrow 7x = 7 \Rightarrow x = 1
$$

Check domain:
- $3x^2 + 1 = 3 + 1 = 4 > 0$ → OK
- $3 + x = 4 > 0$ → OK
- $3x - 2 = 3 - 2 = 1 > 0$ → OK

Valid.

Answer: $x = 1$

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u) $\frac{\log(x^2 + 13)}{\log(x + 5)} = 2$

So:
$$
\log(x^2 + 13) = 2 \log(x + 5) = \log((x + 5)^2)
\Rightarrow x^2 + 13 = (x + 5)^2 = x^2 + 10x + 25
\Rightarrow 13 = 10x + 25 \Rightarrow -12 = 10x \Rightarrow x = -1.2
$$

Check domain:
- $x^2 + 13 > 0$ → always true
- $x + 5 > 0 \Rightarrow x > -5$: $-1.2 > -5$ → OK

But $\log(x + 5)$ is defined at $x = -1.2$: $x+5 = 3.8 > 0$

Now check if logs are defined: yes.

But let’s plug back:
- LHS: $\log((-1.2)^2 + 13) = \log(1.44 + 13) = \log(14.44)$
- RHS: $2 \log(3.8) = \log(3.8^2) = \log(14.44)$ → equal

So valid.

Answer: $x = -1.2 = -\frac{6}{5}$

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v) $\log(3x - 1) - \log(3x + 1) = \log 16$

Use log rule:
$$
\log\left(\frac{3x - 1}{3x + 1}\right) = \log 16
\Rightarrow \frac{3x - 1}{3x + 1} = 16
$$

Now solve:
$$
3x - 1 = 16(3x + 1) = 48x + 16
\Rightarrow 3x - 1 = 48x + 16 \Rightarrow -1 - 16 = 48x - 3x \Rightarrow -17 = 45x \Rightarrow x = -\frac{17}{45}
$$

Check domain:
- $3x - 1 = 3*(-17/45) - 1 = -51/45 - 45/45 = -96/45 < 0$ → invalid!
- Log of negative number → undefined

So no solution

Answer: No solution

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| a) | $x = \sqrt{2}$ |
| b) | $x = 1024$ |
| c) | $x = \pm 10$ |
| d) | $x = \pm 19683$ |
| e) | True (identity) |
| f) | Possibly typo; if $\log_3 x = 7$, then $x = 2187$ |
| g) | $x = 16384$ |
| h) | $x = \frac{1}{81}$ |
| i) | $x = 64$ |
| j) | $x = 4^{1/3}$ |
| k) | $x = 500$ |
| l) | False (no solution) |
| m) | No solution (domain conflict) |
| n) | $x = 3$ |
| o) | $x = 2$ |
| p) | $x = 9$ or $x = -1$ |
| q) | False |
| r) | $x = \frac{3}{4}$ |
| s) | $x = 10^{13/6}$ |
| t) | $x = 1$ |
| u) | $x = -\frac{6}{5}$ |
| v) | No solution |

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