Find the real value of x that satisfies the exponential equation \frac{8^{x + 2}}{4^{x - 1… — JAMB UTME

Find the real value of

x

that satisfies the exponential equation

\frac{8^{x + 2}}{4^{x - 1}} = 16^{x - 1}

Answer: 4

The value of

x

4

Rewriting the terms in base 2 gives

2^{3(x+2)} / 2^{2(x-1)} = 2^{4(x-1)}

. Applying the quotient rule gives an exponent of

(3x + 6) - (2x - 2) = x + 8

on the left. Equating the exponents yields

x + 8 = 4x - 4

, which solves cleanly to

x = 4

Express all terms using a common prime base of 2

The equation becomes

\frac{(2^3)^{x + 2}}{(2^2)^{x - 1}} = (2^4)^{x - 1}

Converting to a common base enables the use of index laws to simplify the equation.

Apply the power-of-a-power and quotient laws of indices

The left side simplifies to

2^{3(x+2) - 2(x-1)} = 2^{x + 8}

and the right side is

2^{4x - 4}

When dividing powers of the same base, exponents are subtracted:

a^m \div a^n = a^{m-n}

Equate exponents and solve the linear equation

x + 8 = 4x - 4 \implies 3x = 12 \implies x = 4

Because the bases on both sides are identical, their respective exponents must be equal.

Solving exponential equations using common base conversion and laws of indices

Question