HONORS ALGEBRA 2 : Problems on Solving Logarithmic Equations

Solve each of the following equations without using calculator.

Problem 1 :

ln x = -1

Solution :

ln x = -1

We know that the base of a natural logarithm is e.

Since ln referes to natural logarithm, its base is e.

ln_e x = -1

Convert to exponential form.

x = e^-1

Problem 2 :

log (x + 3) - 2 = 0

Solution :

log (x + 3) - 2 = 0

Add 2 to both sides.

log (x + 3) = 2

Since the base is not mentioned for log (x + 3), we have to understand that it is a common logarithm. We know that the base of a common logarithm is 10.

Then, we have

log₁₀ (x + 3) = 2

Convert to exponential form.

x + 3 = 10²

x + 3 = 100

Subtract 3 from both sides.

x = 97

Problem 3 :

x = log₁₆ 32

Solution :

x = log₁₆ 32

x = log₁₆ (16 ⋅ 2)

Using the product of rule of logarithms,

x = log₁₆ 16 + log₁₆ 2

x = 1 + log₁₆ 2

Problem 4 :

2log₄ x + 1 = 0

Solution :

2log₄ x + 1 = 0

Subtract 1 from both sides.

2log₄ x = -1

Divide both sides by 2.

Convert to exponential form.

Problem 5 :

Solution :

Multiply both sides byt 4.

log_x 256 = 4

Convert to exponential form.

256 = x⁴

4⁴ = x⁴

4 = x

Problem 6 :

3ln (5x) = 10

Solution :

3ln (5x) = 10

Divide both sides by 3.

Convert to exponential form.

Problem 7 :

log₃ x + log₃ (x - 8) = 2

Solution :

log₃ x + log₃ (x - 8) = 2

Using the product rule of logarithm,

log₃ x(x - 8) = 2

log₃ (x² - 8x) = 2

Convert to exponential form.

x² - 8x = 3²

x² - 8x = 9

x² - 8x - 9 = 0

Solve by factoring.

x² - 9x + x - 9 = 0

x(x - 9) + 1(x - 9) = 0

(x - 9)(x + 1) = 0

x - 9 = 0  or  x + 1 = 0

If x = -1, the arguments of both the logarithms in the given equation become negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = -1.

Therefore,

x = 9

Problem 8 :

log₄ x - log₄ (2x - 3) = 3

Solution :

log₄ x - log₄ (2x - 3) = 3

Using the quotient rule of logarithm,

Convert to exponential form.

x = 64(2x - 3)

x = 128x - 192

192 = 127x

Problem 9 :

3log x - log (x²) = 2

Solution :

3log x - log (x²) = 2

Use the properties of logarithms.

log (x³) - log (x²) = 2

log x = 2

Since the base is not mentioned for log x, we have to understand that it is a common logarithm. We know that the base of a common logarithm is 10.

Then, we have

log₁₀ x = 2

Convert to exponential form.

x = 10²

x = 100

Problem 10 :

2log₆ x + log₆ (x + 1) = log₆ (2x)

Solution :

2log₆ x + log₆ (x + 1) = log₆ (2x)

Use the properties of logarithms.

log₆ (x²) + log₆ (x + 1) = log₆ (2x)

log₆ x²(x + 1) = log₆ (2x)

Since two logarithms are equal with the same base (6), arguments can be equated.

x²(x + 1) = 2x

x³ + x² = 2x

x³ + x² - 2x = 0

x(x² + x - 2) = 0

x = 0  or  x² + x - 2 = 0

Solve  x² + x - 2 = 0 by factoring.

x² + 2x - x - 2 = 0

x(x + 2) - 1(x + 2) = 0

(x + 2)(x - 1) = 0

x + 2 = 0  or  x - 1 = 0

Solving the given equation, we get three values for x.

x = 0, -2, 1

If x = 0 or -2, the arguments of the logarithms in the given equation become zero or negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = 0 and x = -2.

Therefore,

x = 1

Problem 11 :

Solution :

x(x - 6) = 5(x - 2)

x² - 6x = 5x - 10

x² - 11x + 10 = 0

x² - 10x - x + 10 = 0

x(x - 10) - 1(x - 10) = 0

(x - 10)(x - 1) = 0

x - 10 = 0  or  x - 1 = 0

Solving the given equation, we get three values for x.

x = 0, -2, 1

If x = 0 or -2, the arguments of the logarithms in the given equation become zero or negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = 0 and x = -2.

Therefore,

x = 1

If x = 1, the arguments of both the logarithms on the left side of the given equation become negative. Logarithm is defined only for positive values of its argument. So, we can ignore x = 1.

Therefore,

x = 10

Problem 12 :

log (log x) = 1

Solution :

log (log x) = 1

Since the base is not mentioned for the logarithms in the above equation, we have to understand that they are common logarithms. We know that the base of a common logarithm is 10.

Then, we have

log₁₀ (log₁₀ x) = 1

log₁₀ x = 10¹

log₁₀ x = 10

x = 10¹⁰

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