If 8^{x - 1} = 32, what is the value of x? — JAMB UTME

If $8^{x - 1} = 32$ , what is the value of $x$ ?

The value of

x

\frac{8}{3}

To solve

8^{x - 1} = 32

, rewrite both sides using a base of

2

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

. Applying the index law

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

yields

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

. Since the bases are identical, set the exponents equal to each other:

3x - 3 = 5

. Solving for

x

gives

3x = 8

, so

x = \frac{8}{3}

Express both numbers as powers of a common base,

2

8 = 2^3

and

32 = 2^5

, so

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

Exponential equations with different bases are easiest to solve by expressing terms with the same base.

Apply the law of indices

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

to simplify the left-hand side.

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

, which expands to

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

Multiplying the inner exponent by the outer exponent removes parentheses.

Equate the exponents since the bases are equal.

3x - 3 = 5

a^P = a^Q

for

a > 0

and

a \neq 1

, then

P = Q

Solve the linear equation for

x

3x = 5 + 3 \implies 3x = 8 \implies x = \frac{8}{3}

Isolate

x

by adding

3

to both sides and dividing by

3

Solving Exponential Equations using Base Conversion

Tahmini Süre:1m 0s

Soru