What is the value of x that satisfies the exponential equation \frac{4^{2x + 1} \times 8^{… — JAMB UTME

What is the value of

x

that satisfies the exponential equation

\frac{4^{2x + 1} \times 8^{x - 1}}{16^x} = 32

The value of

x

2

Converting all terms to base

2

gives

\frac{2^{4x+2} \times 2^{3x-3}}{2^{4x}} = 2^5

. Combining the powers on the left side yields

2^{3x-1} = 2^5

. Equating exponents gives

3x - 1 = 5

, which solves to

x = 2

Express all base numbers in terms of a common prime base (base 2).

4 = 2^2

8 = 2^3

16 = 2^4

, and

32 = 2^5

Converting all terms to powers of 2 enables the application of index laws.

Substitute these prime base powers into the original equation and expand exponents.

\frac{(2^2)^{2x + 1} \times (2^3)^{x - 1}}{(2^4)^x} = 2^5 \implies \frac{2^{4x + 2} \times 2^{3x - 3}}{2^{4x}} = 2^5

Applying

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

requires multiplying the outer exponent by every term in the inner exponent.

Apply index laws for multiplication (

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

) and division (

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

) on the left-hand side.

2^{(4x + 2) + (3x - 3) - 4x} = 2^5 \implies 2^{3x - 1} = 2^5

Powers with the same base are combined by adding numerator exponents and subtracting denominator exponents.

Equate the exponents since the bases are identical, and solve for

x

3x - 1 = 5 \implies 3x = 6 \implies x = 2

a^x = a^y

for

a > 0

and

a \neq 1

, then

x = y

Solving Exponential Equations using Laws of Indices

Estimated Time:1m 30s

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