What is the sum of all real values of x that satisfy the logarithmic equation \log_3(9^x +… — JAMB UTME

What is the sum of all real values of $x$ that satisfy the logarithmic equation $\log_3(9^x + 81) - x = \log_3 30$ ?

The sum of all real values of

x

satisfying the equation is 4.

Using logarithmic properties,

\log_3(9^x + 81) - x = \log_3 30

simplifies to

\frac{9^x + 81}{3^x} = 30

. Substituting

u = 3^x

produces the quadratic equation

u^2 - 30u + 81 = 0

, which yields

u = 3

and

u = 27

. Solving

3^x = 3

and

3^x = 27

gives

x = 1

and

x = 3

. Their sum is

1 + 3 = 4

Express

x

\log_3(3^x)

and apply the logarithm quotient rule.

\log_3\left(\frac{9^x + 81}{3^x}\right) = \log_3 30

Using

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

combines terms into a single logarithm.

Equate the logarithmic arguments and split the fraction.

\frac{9^x + 81}{3^x} = 30 \implies 3^x + \frac{81}{3^x} = 30

Logarithmic equality with identical bases implies equality of their arguments.

Substitute

u = 3^x

to form and solve a quadratic equation.

u + \frac{81}{u} = 30 \implies u^2 - 30u + 81 = 0 \implies (u - 3)(u - 27) = 0 \implies u = 3 \text{ or } u = 27

Since

9^x = (3^x)^2 = u^2

, substituting

u = 3^x

converts the equation into a standard quadratic equation.

Solve for

x

from

3^x = u

and calculate the sum of the solutions.

3^x = 3 \implies x = 1; \quad 3^x = 27 \implies x = 3. \quad \text{Sum} = 1 + 3 = 4

Matching exponential bases gives

x = 1

and

x = 3

Logarithmic quotient identity and quadratic substitution for exponential equations

Estimated Time:2m 0s

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