If \log_3(x + 5) - \log_{\frac{1}{3}}(x - 1) = 3, find the real value of x. — JAMB UTME

If $\log_3(x + 5) - \log_{\frac{1}{3}}(x - 1) = 3$ , find the real value of $x$ .

Answer: 4

The real value of

x

is 4.

Using the change of base formula,

\log_{\frac{1}{3}}(x - 1) = -\log_3(x - 1)

. Substituting this back into the equation transforms it into

\log_3(x + 5) + \log_3(x - 1) = 3

. Combining the logarithms using the product property gives

\log_3[(x + 5)(x - 1)] = 3

, which means

(x + 5)(x - 1) = 3^3 = 27

. Expanding leads to

x^2 + 4x - 32 = 0

, factoring into

(x + 8)(x - 4) = 0

. Since logarithms require positive arguments (

x > 1

x = -8

is invalid, giving the final answer

x = 4

Apply change of base to express the equation in a single base

\log_{\frac{1}{3}}(x - 1) = \frac{\log_3(x - 1)}{\log_3(1/3)} = -\log_3(x - 1)

Change of base rule

\log_b a = \frac{\log_c a}{\log_c b}

with base

c = 3

, where

\log_3(1/3) = -1

Substitute back into the original equation and combine terms

\log_3(x + 5) - [-\log_3(x - 1)] = \log_3(x + 5) + \log_3(x - 1) = 3

Subtracting a negative logarithm equals adding the positive logarithm.

Apply the product rule of logarithms

\log_3[(x + 5)(x - 1)] = 3

Product rule:

\log_b M + \log_b N = \log_b(MN)

Convert logarithmic equation to quadratic form and solve

(x + 5)(x - 1) = 3^3 = 27 \implies x^2 + 4x - 32 = 0 \implies (x + 8)(x - 4) = 0

Logarithmic definition

\log_b Y = k \implies Y = b^k

Verify domain constraints

x = 4

Logarithmic domain requires

x - 1 > 0 \implies x > 1

. Therefore,

x = -8

is extraneous.

Logarithmic Change of Base and Algebraic Reduction

Question