8+ College Algebra Worksheet Templates - DOC, PDF - Free Printable

Let’s solve each logarithmic equation step-by-step, applying logarithmic properties and algebraic techniques.

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1. \( 2 \log_4(x) + 3 = 0 \)

Step 1: Isolate the log term.
\[
2 \log_4(x) = -3
\]
\[
\log_4(x) = -\frac{3}{2}
\]

Step 2: Convert to exponential form.
\[
x = 4^{-3/2} = (2^2)^{-3/2} = 2^{-3} = \frac{1}{8}
\]

✔ Answer: \( \boxed{\dfrac{1}{8}} \)

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2. \( \log_4 x + \log_4(2x + 1) = 0 \)

Step 1: Use product rule: \( \log_b A + \log_b B = \log_b(AB) \)
\[
\log_4 [x(2x + 1)] = 0
\]

Step 2: Convert to exponential form.
\[
x(2x + 1) = 4^0 = 1
\]
\[
2x^2 + x - 1 = 0
\]

Step 3: Solve quadratic.
\[
x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}
\Rightarrow x = \frac{1}{2}, \quad x = -1
\]

Step 4: Check domain — logs require positive arguments.

- For \( x = \frac{1}{2} \): \( x > 0 \), \( 2x+1 = 2 > 0 \) → OK.
- For \( x = -1 \): \( x < 0 \) → INVALID.

✔ Answer: \( \boxed{\dfrac{1}{2}} \)

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3. \( \log x - \log(x^2 - 1) = -2 \log(x - 1) \)

Assume base 10 (common log). Also note: \( \log(x^2 - 1) = \log[(x-1)(x+1)] \)

Step 1: Apply quotient rule on left:
\[
\log\left( \frac{x}{x^2 - 1} \right) = -2 \log(x - 1)
\]

Step 2: Use power rule on right: \( -2 \log(x - 1) = \log\left( (x - 1)^{-2} \right) \)

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

Step 3: Since logs are equal, arguments must be equal (if same base and defined):
\[
\frac{x}{x^2 - 1} = \frac{1}{(x - 1)^2}
\]

Note: \( x^2 - 1 = (x - 1)(x + 1) \), so:

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

Multiply both sides by \( (x - 1)^2(x + 1) \) (assuming \( x \neq 1, -1 \)):

\[
x(x - 1) = x + 1
\]
\[
x^2 - x = x + 1
\]
\[
x^2 - 2x - 1 = 0
\]

Step 4: Solve quadratic.
\[
x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}
\]

Step 5: Check domain — all logs must have positive arguments.

Recall: \( x > 0 \), \( x^2 - 1 > 0 \Rightarrow |x| > 1 \), and \( x - 1 > 0 \Rightarrow x > 1 \).

So we need \( x > 1 \).

- \( 1 + \sqrt{2} \approx 2.414 > 1 \) → OK.
- \( 1 - \sqrt{2} \approx -0.414 < 1 \) → INVALID.

✔ Answer: \( \boxed{1 + \sqrt{2}} \)

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4. \( 2 \ln(x + 3) - \ln(x + 1) = 3 \ln 2 \)

Use natural log (base e).

Step 1: Apply power rule: \( 2 \ln(x+3) = \ln((x+3)^2) \), \( 3 \ln 2 = \ln(8) \)

So:
\[
\ln\left( \frac{(x + 3)^2}{x + 1} \right) = \ln(8)
\]

Step 2: Since ln is one-to-one:
\[
\frac{(x + 3)^2}{x + 1} = 8
\]

Step 3: Multiply both sides by \( x + 1 \) (assume \( x \ne -1 \)):
\[
(x + 3)^2 = 8(x + 1)
\]
\[
x^2 + 6x + 9 = 8x + 8
\]
\[
x^2 - 2x + 1 = 0
\]
\[
(x - 1)^2 = 0 \Rightarrow x = 1
\]

Step 4: Check domain: \( x + 3 > 0 \), \( x + 1 > 0 \) → \( x > -1 \). So \( x = 1 \) is valid.

✔ Answer: \( \boxed{1} \)

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5. \( \log_2(2x - 9) = 2 - \log_2(x - 1) \)

Step 1: Move all logs to one side:
\[
\log_2(2x - 9) + \log_2(x - 1) = 2
\]

Step 2: Use product rule:
\[
\log_2[(2x - 9)(x - 1)] = 2
\]

Step 3: Convert to exponential form:
\[
(2x - 9)(x - 1) = 2^2 = 4
\]
\[
2x^2 - 2x - 9x + 9 = 4
\]
\[
2x^2 - 11x + 5 = 0
\]

Step 4: Solve quadratic.
\[
x = \frac{11 \pm \sqrt{121 - 40}}{4} = \frac{11 \pm \sqrt{81}}{4} = \frac{11 \pm 9}{4}
\Rightarrow x = 5, \quad x = \frac{1}{2}
\]

Step 5: Check domain — arguments of logs must be positive.

- \( 2x - 9 > 0 \Rightarrow x > 4.5 \)
- \( x - 1 > 0 \Rightarrow x > 1 \)

So only \( x = 5 \) satisfies.

✔ Answer: \( \boxed{5} \)

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6. \( \log(x - 2)(2x - 3) = \log x^2 \)

This is ambiguous — does it mean \( \log[(x - 2)(2x - 3)] = \log(x^2) \)?
Yes, that’s the standard interpretation.

Step 1: Since logs are equal, arguments are equal (if defined):
\[
(x - 2)(2x - 3) = x^2
\]

Step 2: Expand left:
\[
2x^2 - 3x - 4x + 6 = x^2
\]
\[
2x^2 - 7x + 6 = x^2
\]
\[
x^2 - 7x + 6 = 0
\]
\[
(x - 1)(x - 6) = 0 \Rightarrow x = 1, \quad x = 6
\]

Step 3: Check domain — all arguments > 0.

- For \( x = 1 \): \( x - 2 = -1 < 0 \), \( 2x - 3 = -1 < 0 \), \( x^2 = 1 > 0 \) → INVALID (negative inside log).
- For \( x = 6 \): \( x - 2 = 4 > 0 \), \( 2x - 3 = 9 > 0 \), \( x^2 = 36 > 0 \) → VALID.

✔ Answer: \( \boxed{6} \)

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7. \( \log_x\left(\frac{1}{9}\right) = 2 \)

Step 1: Convert to exponential form:
\[
x^2 = \frac{1}{9}
\]

Step 2: Solve for x:
\[
x = \pm \frac{1}{3}
\]

Step 3: Check domain — base of log must be positive and ≠ 1.

- \( x = \frac{1}{3} > 0 \), ≠ 1 → OK.
- \( x = -\frac{1}{3} < 0 \) → INVALID.

✔ Answer: \( \boxed{\dfrac{1}{3}} \)

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## ✔ Final Answers:

1. \( \boxed{\dfrac{1}{8}} \)
2. \( \boxed{\dfrac{1}{2}} \)
3. \( \boxed{1 + \sqrt{2}} \)
4. \( \boxed{1} \)
5. \( \boxed{5} \)
6. \( \boxed{6} \)
7. \( \boxed{\dfrac{1}{3}} \)