SOLVING LOGARITHMIC EQUATIONS BY EXPONENTIATING

Solved each of the following logarithmic equations by exponentiating.

Problem 1 :

log₄(x - 5) = 3

Solution :

Since the base of the logarithm is 4, exponentiate both sides of the equation with base 4.

x - 5 = 4³

x - 5 = 64

Add 5 to both sides.

x = 69

Problem 2 :

log(5x) = log(2x + 9)

Solution :

Since the base is not given for both the logarithms, we have to assume that they are common logarithms. Hence the base for both the logarithms is 10.

log₁₀(5x) = log₁₀(2x + 9)

Since the base of the logarithms is 10, exponentiate both sides of the equation with base 10.

5x = 2x + 9

Subtract 2x from both sides.

3x = 9

Divide both sides by 3.

x = 3

Problem 3 :

ln(5x - 1) = ln(7x - 9)

Solution :

Since the given logarithms are natural logarithms, their base is e.

So, exponentiate both sides of the equation with base e.

5x - 1 = 7x - 9

Subtract 7x from both sides.

-2x - 1 = -9

Add 1 to both sides.

-2x = -8

Divide both sides by (-2).

x = 4

Problem 4 :

Solution :

The bases of the logarithms are different, they are 3 and 9.

Here, 3 can not be written as a power of 9.

But 9 can be written as a power of 3, that is

9 = 3²

So, exponentiate both sides of the equation with base 9.

3x + 7 = 2x + 10

Subtract 2x from both sides.

x + 7 = 10

Subtract 7 from both sides.

x = 3

Problem 5 :

log₁₄(x - 3) + log₁₄(x + 2) = 1

Solution :

log₁₄(x - 3) + log₁₄(x + 2) = 1

log₁₄(x - 3) + log₁₄(x + 2) = log₁₄14

log₁₄[(x - 3)(x + 2)] = log₁₄14

log₁₄(x² - x - 6) = log₁₄14

Since the base of the logarithms is 14, exponentiate both sides of the equation with base 14.

x² - x - 6 = 14

Subtract 14 from both sides.

x² - x - 20 = 0

Solve by factoring.

x² - 5x + 4x - 20 = 0

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

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

The aruguments of logarithms are always positive or greater than zero.

The common solution of the inequalities x > 3 and x > -2 is

x > 3

Since x > 3, the solution x = -4 can not be acccepted.

Therefore,

x = 5

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