Find the positive integer value of x that satisfies the logarithmic equation x^{\log_3 x} … — JAMB UTME

Find the positive integer value of $x$ that satisfies the logarithmic equation $x^{\log_3 x} = 81x^3$ .

Answer: 81

The positive integer value of

x

is 81.

Taking the logarithm base 3 on both sides transforms

x^{\log_3 x} = 81x^3

into

(\log_3 x)^2 = 4 + 3\log_3 x

. Substituting

u = \log_3 x

yields the quadratic equation

u^2 - 3u - 4 = 0

, which factors as

(u-4)(u+1) = 0

. The roots are

u = 4

(

x = 3^4 = 81

) and

u = -1

(

x = 3^{-1} = 1/3

). Since

x

must be a positive integer, the correct value is 81.

Take logarithm base 3 of both sides of the equation

x^{\log_3 x} = 81x^3

\log_3(x^{\log_3 x}) = \log_3(81x^3)

Taking the logarithm with base 3 allows us to simplify the exponent containing

\log_3 x

Apply logarithmic identities

\log_b(a^k) = k \log_b a

and

\log_b(mn) = \log_b m + \log_b n

(\log_3 x)^2 = \log_3 81 + 3\log_3 x = 4 + 3\log_3 x

Expanding the products and powers reduces the equation into a single logarithmic variable

\log_3 x

Substitute

u = \log_3 x

to create a quadratic equation

u^2 - 3u - 4 = 0

Setting

u = \log_3 x

converts the equation into standard quadratic form.

Factor the quadratic expression

(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1

Factoring allows us to find all possible real values for

u

Convert back to

x

using

x = 3^u

and select the positive integer root

x = 3^4 = 81

x = 3^{-1} = \frac{1}{3}

. The positive integer solution is

x = 81

The question specifically requests the positive integer solution, eliminating

x = \frac{1}{3}

Solving equations with variable exponents by taking logarithms and reducing to a quadratic form.

Question