# 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.

lne x = -1

Convert to exponential form.

x = e-1

Problem 2 :

log (x + 3) - 2 = 0

Solution :

log (x + 3) - 2 = 0

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.

logx 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

 x - 9 = 0x = 9 x + 1 = 0x = -1

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

 x + 2 = 0x = -2 x - 1 = 0x = 1

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

 x - 10 = 0x = 10 x - 1 = 0x = 1

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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