Question

Difficulty: MediumSurds and Indices

If 2x+1+2x−1=3202^{x+1} + 2^{x-1} = 320, what is the value of xx?

Answer: 7

Answer

The value of xx is 7.
Factoring out 2x−12^{x-1} from the expression gives 2x−1(22+1)=3202^{x-1}(2^2 + 1) = 320, which simplifies to 5⋅2x−1=3205 \cdot 2^{x-1} = 320. Dividing by 5 yields 2x−1=64=262^{x-1} = 64 = 2^6. Equating exponents gives x−1=6x - 1 = 6, so x=7x = 7.

Step-by-Step Solution

1
Rewrite terms using exponent rules to factor out the common power 2x−12^{x-1}
2x−1(22+1)=3202^{x-1}(2^2 + 1) = 320
Using the product rule am+n=am⋅ana^{m+n} = a^m \cdot a^n, we have 2x+1=2x−1⋅222^{x+1} = 2^{x-1} \cdot 2^2.
2
Simplify the numerical factor inside the parentheses
5⋅2x−1=3205 \cdot 2^{x-1} = 320
22+1=4+1=52^2 + 1 = 4 + 1 = 5.
3
Isolate the exponential term by dividing by 5
2x−1=642^{x-1} = 64
Dividing 320320 by 55 gives 6464.
4
Write 64 as a power of 2 and solve for xx
2x−1=26  ⟹  x−1=6  ⟹  x=72^{x-1} = 2^6 \implies x - 1 = 6 \implies x = 7
Equating exponents when bases are equal.

Key Concept

Solving Exponential Equations by Factoring Common Powers
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