Find the value of x that satisfies the exponential equation 27^{x - 1} = \frac{9^{x + 1}}{… — JAMB UTME

Find the value of $x$ that satisfies the exponential equation $27^{x - 1} = \frac{9^{x + 1}}{3^{x - 5}}$ .

Answer: 5

The value of

x

is 5.

Rewriting the terms with a common base of 3 transforms the equation into

3^{3x-3} = 3^{x+7}

. Equating exponents gives

3x - 3 = x + 7

, which yields

x = 5

Convert all terms to base 3

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

and

9^{x+1} = 3^{2x+2}

Laws of indices require identical bases to manipulate exponents.

Apply division rule of indices to the right-hand side

\frac{3^{2x+2}}{3^{x-5}} = 3^{x+7}

When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.

Equate the exponents and solve for

x

3x - 3 = x + 7 \implies 2x = 10 \implies x = 5

a^f(x) = a^g(x)

for

a > 0

and

a \neq 1

, then

f(x) = g(x)

Solving exponential equations using base reduction and exponent laws

Estimated Time:1m 30s

Question