If \log_3 x - 2\log_x 27 = 1, what is the sum of all possible real values of x? — JAMB UTME

If $\log_3 x - 2\log_x 27 = 1$ , what is the sum of all possible real values of $x$ ?

\frac{244}{9}

By applying the change of base rule

\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}

, the given equation simplifies to

\log_3 x - \frac{6}{\log_3 x} = 1

. Setting

u = \log_3 x

yields

u^2 - u - 6 = 0

, which factors as

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

. Thus,

u = 3

u = -2

, giving solutions

x = 3^3 = 27

and

x = 3^{-2} = \frac{1}{9}

. Adding these valid real solutions gives

27 + \frac{1}{9} = \frac{244}{9}

Apply the change of base formula to

\log_x 27

\log_x 27 = \frac{\log_3 27}{\log_3 x} = \frac{3}{\log_3 x}

Converting all logarithmic terms to a common base (base 3) allows substitution.

Substitute

\log_x 27

back into the original equation and let

u = \log_3 x

u - 2\left(\frac{3}{u}\right) = 1 \implies u - \frac{6}{u} = 1

This transforms the logarithmic equation into an algebraic equation.

Clear the denominator by multiplying through by

u

and rearrange into standard quadratic form.

u^2 - u - 6 = 0

Multiplying by

u

(where

u \neq 0

) forms a standard quadratic equation.

Factor the quadratic equation to solve for

u

(u - 3)(u + 2) = 0 \implies u = 3 \text{ or } u = -2

Factoring determines the values of

\log_3 x

Convert back to

x

using

x = 3^u

For

u = 3

x = 3^3 = 27

. For

u = -2

x = 3^{-2} = \frac{1}{9}

. Both

x = 27

and

x = \frac{1}{9}

are valid bases (

x > 0, x \neq 1

Exponentiation resolves the logarithmic variable.

Find the sum of all valid values of

x

\text{Sum} = 27 + \frac{1}{9} = \frac{243 + 1}{9} = \frac{244}{9}

Combines the two real solutions into the final requested sum.

Change of base formula

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

and solving equations reducible to quadratics.

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