If 2^{x-1} + 2^{x+1} = 160, what is the value of 3^{x-2}? — JAMB UTME

If $2^{x-1} + 2^{x+1} = 160$ , what is the value of $3^{x-2}$ ?

Factoring

2^{x-1}

from the left-hand side of

2^{x-1} + 2^{x+1} = 160

yields

2^{x-1}(1 + 2^2) = 160

, which reduces to

5 \cdot 2^{x-1} = 160

. Dividing by

5

gives

2^{x-1} = 32 = 2^5

. Equating exponents gives

x - 1 = 5

, so

x = 6

. Substituting

x = 6

into

3^{x-2}

gives

3^{6-2} = 3^4 = 81

Factor out the common term

2^{x-1}

from the expression

2^{x-1} + 2^{x+1}

2^{x-1}(1 + 2^2) = 160

Applying index law

2^{x+1} = 2^{x-1} \cdot 2^2

allows factoring out

2^{x-1}

Simplify the bracketed terms and solve for

2^{x-1}

5 \cdot 2^{x-1} = 160 \implies 2^{x-1} = 32

Dividing both sides by

5

isolates the term with the unknown exponent.

Express

32

as a power of

2

and solve for

x

2^{x-1} = 2^5 \implies x - 1 = 5 \implies x = 6

Equating exponents since the bases are identical.

Substitute

x = 6

into the target expression

3^{x-2}

3^{6-2} = 3^4 = 81

Evaluating the power gives the final answer.

Factoring exponential expressions with common bases and applying laws of indices

Estimated Time:1m 30s

Question