What is the value of x that satisfies the equation 16^{x - 1} \times 4^{x + 2} = 8^{x + 3}… — JAMB UTME

What is the value of $x$ that satisfies the equation $16^{x - 1} \times 4^{x + 2} = 8^{x + 3}$ ?

The value of

x

3

Converting all terms to base 2 gives

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

. Expanding the exponents gives

2^{4x-4} \times 2^{2x+4} = 2^{3x+9}

, which simplifies to

2^{6x} = 2^{3x+9}

. Equating exponents yields

6x = 3x + 9

, giving the solution

x = 3

Express all terms with a common base of 2.

16 = 2^4

4 = 2^2

, and

8 = 2^3

, so the equation becomes

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

Converting all terms to a common prime base allows application of index laws.

Apply power of a power law

(a^m)^n = a^{m n}

2^{4x - 4} \times 2^{2x + 4} = 2^{3x + 9}

Multiply exponents when raising a power to another power.

Apply multiplication law

a^m \times a^n = a^{m+n}

on the left side.

2^{(4x - 4) + (2x + 4)} = 2^{6x}

Add exponents when multiplying powers with the same base.

Equate exponents of equal bases.

6x = 3x + 9 \implies 3x = 9 \implies x = 3

a^m = a^n

for

a > 0

and

a \neq 1

, then

m = n

Laws of Indices and Solving Exponential Equations

Soru