HONORS ALGEBRA 2 : Solving Logarithmic Equations

Problems 1-4 : Solve the given exponential equation and show work leading to exact answer.

Problem 1 :

log(z - 3) = 2

Solution :

log(z - 3) = 2

In the equation above, since the base of the logarithm is not mentioned, we have to understand that it is a common logarithm and base of a common logarithm is 10.

log₁₀(z - 3) = 2

Convert the above exponential equation.

z - 3 = 10²

z - 3 = 100

Add 3 to both sides.

z = 103

Problem 2 :

3ln 5x = 9

Solution :

3ln 5x = 9

Divide both sides by 3.

ln 5x = 3

In the equation above, ln refers to natural logarithm and its base is always e.

ln_e 5x = 3

Convert the above exponential equation.

5x = e³

Divide both sides by 5.

Problem 3 :

log x - log (x - 1) = 1

Solution :

log x - log (x - 1) = 1

Use the quotient rule of logarithm.

Multiply both sides by (x - 1).

x = 10(x - 1)

x = 10x - 10

9x = 10

Problem 4 :

log₃ x + log₃ (x - 8) = 2

Solution :

log₃ x + log₃ (x - 8) = 2

Use the product rule of logarithm.

log₃ [x(x - 8)] = 2

x(x - 8) = 3²

x² - 8x = 9

Subtract 9 from both sides.

x² - 8x - 9 = 0

Solve by factoring.

x² - 9x + x - 9 = 0

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

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

x + 1 = 0  or  x - 9 = 0

x = -1  or x = 9

A logarithm is defined only for the positive value of its argument.

When x = -1, the arguments of the logarithms in the given equation become negative. So, x = -1 can not be accepted.

Therefore,

x = 9

Problem 5 :

Solution :

Problem 6 :

Solution :

Problem 7 :

2log x = log (6 - x)

Solution :

2log x = log (6 - x)

log x² = log (6 - x)

x² = 6 - x

x² + x - 6 = 0

Solve by factoring.

x² + 3x - 2x - 6 = 0

x(x + 3) - 2(x + 3) = 0

(x + 3)(x - 2) = 0

x + 3 = 0  or  x - 2 = 0

x = -3  or  x = 2

We already know that a logarithm is defined only for the positive value of its argument.

When x = -3, the argument of log x in the given equation becomes negative. So, x = -3 can not be accepted.

Therefore,

x = 2

Problem 8 :

log 5x + log (x - 1) = 2

Solution :

log 5x + log (x - 1) = 2

Use the product rule of logarithm.

log [5x(x - 1)] = 2

log₁₀ [5x(x - 1)] = 2

Convert the above equation to exponential form.

5x(x - 1) = 10²

5x² - 5x = 100

Subtract 100 from both sides.

5x² - 5x - 100 = 0

Divide both sides by 5.

x² - x - 20 = 0

Solve by factoring.

x² - 5x + 4x - 20 = 0

x(x - 5) + 4(x - 5) = 0

(x - 5)(x - 4) = 0

x - 5 = 0  or  x - 4 = 0

x = 5  or x = 4

Even though the arguments of the logarithms in the given equation become positive, when x = 4 and x = 5, we have to verify the two solutions x = 4 and x = 5 with the given equations. Because, the equation contains logarithms.

Therefore,

x = 5

Problems 9-10 : Solve the given logarithmic equation and round your answer to three decimal places.

Problem 9 :

x = log₁₅ 32

Solution :

x = log₁₅ 32

x = 1.280

Problem 10 :

ln (5 + 2x) = -0.75

Solution :

ln (5 + 2x) = -0.75

Convert the above equation to exponential form.

5 + 2x = e^-0.75

x = -2.264

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