Logarithm Practice Problems

1. Using the logarithm quotient rule: log_a(x) - log_a(y) = log_a(x/y)

2. Rewrite the equation: log₂((x² - 7) / (x - 3)) = 3

3. Apply 2³ to both sides: (x² - 7) / (x - 3) = 2³ = 8

4. Multiply both sides by (x - 3): x² - 7 = 8(x - 3)

5. Expand: x² - 7 = 8x - 24

6. Rearrange: x² - 8x + 17 = 0

7. This is a quadratic equation. Use the quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

8. a = 1, b = -8, c = 17

9. x = (8 ± √(64 - 68)) / 2 = (8 ± √-4) / 2

10. Since √-4 is not a real number, this equation has no real solutions.

Using the logarithm product rule: log_a(x) + log_a(y) = log_a(xy)

log₃(27) + log₃(9) = log₃(27 * 9) = log₃(243)

log₃(243) = log₃(3⁵) = 5

Using the change of base formula: log₅(125) = log₁₀(125) / log₁₀(5)

= 2.0969 / 0.6990 ≈ 3

We can verify this: 5³ = 125, so log₅(125) = 3

2⁴ = x + 3

16 = x + 3

x = 13

Using the logarithm quotient rule: log_a(x) - log_a(y) = log_a(x/y)

log₁₀(1000) - log₁₀(100) = log₁₀(1000/100) = log₁₀(10)

log₁₀(10) = 1

1. Using the change of base formula: log_a(64) = log₂(64) / log₂(a)

2. Simplify log₂(64): log₂(64) = log₂(2⁶) = 6

3. Substitute this back into the equation:

log_a(64) = 6 / log₂(a)

This expresses log_a(64) in terms of log₂(a).

Using the logarithm quotient rule: log_a(x) - log_a(y) = log_a(x/y)

log₇(49x²) - log₇(x) = log₇(49x² / x) = log₇(49x)

log₇(49x) = log₇(7²x) = log₇(7²) + log₇(x) = 2 + log₇(x)

Start with y = log₃(x)

Apply 3^y to both sides: 3^y = 3^log₃(x)

Simplify the right side: 3^y = x

Therefore, x = 3^y

Using the logarithm product rule: log₂(x-1) + log₂(x+1) = log₂((x-1)(x+1)) = 3

log₂(x² - 1) = 3

x² - 1 = 2³ = 8

x² = 9

x = ±3

Since log(x-1) is only defined for x > 1, x = 3 is the only valid solution.

Start with y = 2^x + 1

Subtract 1 from both sides: y - 1 = 2^x

Take the logarithm (base 2) of both sides: log₂(y - 1) = x

Therefore, x = log₂(y - 1)