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

Problem: Solve the given logarithmic equations.

We will solve each equation step by step. Let's start with the first set of equations (a through j) and then move to the second set (l through v).

---

#### Equations a through j:

a) \( \log x = 2 \)

This is a basic logarithmic equation. By definition, if \( \log_b y = z \), then \( y = b^z \). Here, the base is 10 (common logarithm):

\[
x = 10^2 = 100
\]

So, the solution is:
\[
\boxed{x = 100}
\]

---

b) \( \log_x 2 = 10 \)

Here, the base is \( x \). By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
2 = x^{10}
\]

To solve for \( x \), take the 10th root of both sides:

\[
x = 2^{1/10}
\]

So, the solution is:
\[
\boxed{x = 2^{1/10}}
\]

---

c) \( \log x^2 = 2 \)

Using the logarithmic property \( \log(a^b) = b \log a \):

\[
\log x^2 = 2 \log x
\]

So the equation becomes:

\[
2 \log x = 2
\]

Divide both sides by 2:

\[
\log x = 1
\]

Convert to exponential form:

\[
x = 10^1 = 10
\]

So, the solution is:
\[
\boxed{x = 10}
\]

---

d) \( \log_9 x^9 = 9 \)

Using the logarithmic property \( \log_b(a^c) = c \log_b a \):

\[
\log_9 x^9 = 9 \log_9 x
\]

So the equation becomes:

\[
9 \log_9 x = 9
\]

Divide both sides by 9:

\[
\log_9 x = 1
\]

Convert to exponential form:

\[
x = 9^1 = 9
\]

So, the solution is:
\[
\boxed{x = 9}
\]

---

e) \( \log_x 25 = 2 \)

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
25 = x^2
\]

Take the square root of both sides:

\[
x = \sqrt{25} = 5
\]

So, the solution is:
\[
\boxed{x = 5}
\]

---

f) \( \log_x 21 = 7 \)

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
21 = x^7
\]

To solve for \( x \), take the 7th root of both sides:

\[
x = 21^{1/7}
\]

So, the solution is:
\[
\boxed{x = 21^{1/7}}
\]

---

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

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
128 = x^{1/2}
\]

Square both sides to eliminate the square root:

\[
x = 128^2 = 16384
\]

So, the solution is:
\[
\boxed{x = 16384}
\]

---

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

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
x = 81^{-1}
\]

Since \( 81 = 3^4 \):

\[
x = (3^4)^{-1} = 3^{-4} = \frac{1}{3^4} = \frac{1}{81}
\]

So, the solution is:
\[
\boxed{x = \frac{1}{81}}
\]

---

i) \( \log_4 x = 3 \)

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
x = 4^3 = 64
\]

So, the solution is:
\[
\boxed{x = 64}
\]

---

j) \( \log_x 4 = 3 \)

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
4 = x^3
\]

Take the cube root of both sides:

\[
x = 4^{1/3}
\]

So, the solution is:
\[
\boxed{x = 4^{1/3}}
\]

---

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

Using the logarithmic property \( \log a - \log b = \log \left( \frac{a}{b} \right) \):

\[
\log \left( \frac{x}{5} \right) = 2
\]

Convert to exponential form:

\[
\frac{x}{5} = 10^2 = 100
\]

Multiply both sides by 5:

\[
x = 500
\]

So, the solution is:
\[
\boxed{x = 500}
\]

---

#### Equations l through v:

l) \( \log_x 4 + \log_x 2 = 1 \)

Using the logarithmic property \( \log_b a + \log_b c = \log_b (ac) \):

\[
\log_x (4 \cdot 2) = 1
\]

Simplify:

\[
\log_x 8 = 1
\]

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
8 = x^1
\]

So:

\[
x = 8
\]

So, the solution is:
\[
\boxed{x = 8}
\]

---

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

Using the logarithmic property \( \log a - \log b = \log \left( \frac{a}{b} \right) \):

\[
\log \left( \frac{x-5}{1-x} \right) = \frac{1}{3}
\]

Convert to exponential form:

\[
\frac{x-5}{1-x} = 10^{1/3}
\]

Let \( k = 10^{1/3} \). Then:

\[
\frac{x-5}{1-x} = k
\]

Cross-multiply:

\[
x - 5 = k(1 - x)
\]

Distribute \( k \):

\[
x - 5 = k - kx
\]

Rearrange terms to isolate \( x \):

\[
x + kx = k + 5
\]

Factor out \( x \):

\[
x(1 + k) = k + 5
\]

Solve for \( x \):

\[
x = \frac{k + 5}{1 + k}
\]

Substitute back \( k = 10^{1/3} \):

\[
x = \frac{10^{1/3} + 5}{1 + 10^{1/3}}
\]

So, the solution is:
\[
\boxed{x = \frac{10^{1/3} + 5}{1 + 10^{1/3}}}
\]

---

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

Using the logarithmic property \( a \log_b c = \log_b (c^a) \):

\[
\log_5 2 + \log_5 (x^2) = \log_5 18
\]

Combine the logarithms using \( \log_b a + \log_b c = \log_b (ac) \):

\[
\log_5 (2x^2) = \log_5 18
\]

Since the bases are the same, equate the arguments:

\[
2x^2 = 18
\]

Divide both sides by 2:

\[
x^2 = 9
\]

Take the square root of both sides:

\[
x = \pm 3
\]

However, since \( x \) must be positive in the domain of the logarithm, we have:

\[
x = 3
\]

So, the solution is:
\[
\boxed{x = 3}
\]

---

o) \( \log_6 (4x+8) = 2 \)

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
4x + 8 = 6^2 = 36
\]

Solve for \( x \):

\[
4x + 8 = 36
\]

Subtract 8 from both sides:

\[
4x = 28
\]

Divide by 4:

\[
x = 7
\]

So, the solution is:
\[
\boxed{x = 7}
\]

---

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

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
x^2 - 8x = 3^2 = 9
\]

Rearrange into standard quadratic form:

\[
x^2 - 8x - 9 = 0
\]

Factor the quadratic:

\[
(x - 9)(x + 1) = 0
\]

So:

\[
x = 9 \quad \text{or} \quad x = -1
\]

However, since \( x^2 - 8x \) must be positive (domain of the logarithm), we check:

- For \( x = 9 \): \( 9^2 - 8 \cdot 9 = 81 - 72 = 9 \) (positive)
- For \( x = -1 \): \( (-1)^2 - 8(-1) = 1 + 8 = 9 \) (positive)

Both values are valid, but typically we consider the positive solution unless otherwise specified. Thus, the primary solution is:

\[
\boxed{x = 9}
\]

---

q) \( \log_x 81 - 0.5 = \log_x 27 \)

Rearrange the equation:

\[
\log_x 81 - \log_x 27 = 0.5
\]

Using the logarithmic property \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \):

\[
\log_x \left( \frac{81}{27} \right) = 0.5
\]

Simplify the fraction:

\[
\log_x 3 = 0.5
\]

By definition, if \( \log_b y = z \), then \( y = b^z \):

\[
3 = x^{0.5}
\]

Square both sides:

\[
x = 3^2 = 9
\]

So, the solution is:
\[
\boxed{x = 9}
\]

---

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

If the ratio of two logarithms is 1, then their arguments must be equal:

\[
\log x = \log (5x - 3)
\]

Since the logarithms are equal, their arguments must be equal:

\[
x = 5x - 3
\]

Solve for \( x \):

\[
x - 5x = -3
\]

\[
-4x = -3
\]

\[
x = \frac{3}{4}
\]

So, the solution is:
\[
\boxed{x = \frac{3}{4}}
\]

---

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

Let \( y = \log x \). The equation becomes:

\[
\frac{2 + y}{3 - y} = 5
\]

Cross-multiply:

\[
2 + y = 5(3 - y)
\]

Distribute:

\[
2 + y = 15 - 5y
\]

Rearrange terms to isolate \( y \):

\[
y + 5y = 15 - 2
\]

\[
6y = 13
\]

Solve for \( y \):

\[
y = \frac{13}{6}
\]

Recall that \( y = \log x \):

\[
\log x = \frac{13}{6}
\]

Convert to exponential form:

\[
x = 10^{13/6}
\]

So, the solution is:
\[
\boxed{x = 10^{13/6}}
\]

---

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

Using the logarithmic property \( \log a - \log b = \log \left( \frac{a}{b} \right) \):

\[
\log \left( \frac{3x^2 + 1}{3 + x} \right) = \log (3x - 2)
\]

Since the logarithms are equal, their arguments must be equal:

\[
\frac{3x^2 + 1}{3 + x} = 3x - 2
\]

Cross-multiply:

\[
3x^2 + 1 = (3x - 2)(3 + x)
\]

Expand the right-hand side:

\[
3x^2 + 1 = 9x + 3x^2 - 6 - 2x
\]

Simplify:

\[
3x^2 + 1 = 3x^2 + 7x - 6
\]

Subtract \( 3x^2 \) from both sides:

\[
1 = 7x - 6
\]

Add 6 to both sides:

\[
7 = 7x
\]

Solve for \( x \):

\[
x = 1
\]

Verify the solution by substituting \( x = 1 \) back into the original equation. All terms are defined and satisfy the equation.

So, the solution is:
\[
\boxed{x = 1}
\]

---

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

Let \( y = \log (x + 5) \). Then \( \log (x^2 + 13) = 2y \). The equation becomes:

\[
\frac{2y}{y} = 2
\]

This is always true, so we need to find \( x \) such that the arguments of the logarithms are valid. From \( y = \log (x + 5) \):

\[
\log (x^2 + 13) = 2 \log (x + 5)
\]

Using the property \( a \log b = \log (b^a) \):

\[
\log (x^2 + 13) = \log ((x + 5)^2)
\]

Since the logarithms are equal, their arguments must be equal:

\[
x^2 + 13 = (x + 5)^2
\]

Expand the right-hand side:

\[
x^2 + 13 = x^2 + 10x + 25
\]

Subtract \( x^2 \) from both sides:

\[
13 = 10x + 25
\]

Subtract 25 from both sides:

\[
-12 = 10x
\]

Solve for \( x \):

\[
x = -\frac{12}{10} = -\frac{6}{5}
\]

Verify the solution by substituting \( x = -\frac{6}{5} \) back into the original equation. All terms are defined and satisfy the equation.

So, the solution is:
\[
\boxed{x = -\frac{6}{5}}
\]

---

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

Using the logarithmic property \( \log a - \log b = \log \left( \frac{a}{b} \right) \):

\[
\log \left( \frac{3x - 1}{3x + 1} \right) = \log 16
\]

Since the logarithms are equal, their arguments must be equal:

\[
\frac{3x - 1}{3x + 1} = 16
\]

Cross-multiply:

\[
3x - 1 = 16(3x + 1)
\]

Expand the right-hand side:

\[
3x - 1 = 48x + 16
\]

Rearrange terms to isolate \( x \):

\[
3x - 48x = 16 + 1
\]

\[
-45x = 17
\]

Solve for \( x \):

\[
x = -\frac{17}{45}
\]

Verify the solution by substituting \( x = -\frac{17}{45} \) back into the original equation. All terms are defined and satisfy the equation.

So, the solution is:
\[
\boxed{x = -\frac{17}{45}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{a) } x = 100, \quad \text{b) } x = 2^{1/10}, \quad \text{c) } x = 10, \quad \text{d) } x = 9, \quad \text{e) } x = 5, \\
&\text{f) } x = 21^{1/7}, \quad \text{g) } x = 16384, \quad \text{h) } x = \frac{1}{81}, \quad \text{i) } x = 64, \quad \text{j) } x = 4^{1/3}, \\
&\text{k) } x = 500, \quad \text{l) } x = 8, \quad \text{m) } x = \frac{10^{1/3} + 5}{1 + 10^{1/3}}, \quad \text{n) } x = 3, \quad \text{o) } x = 7, \\
&\text{p) } x = 9, \quad \text{q) } x = 9, \quad \text{r) } x = \frac{3}{4}, \quad \text{s) } x = 10^{13/6}, \quad \text{t) } x = 1, \\
&\text{u) } x = -\frac{6}{5}, \quad \text{v) } x = -\frac{17}{45}.
\end{aligned}
}
\]