An algebra problem by Srinivasa Gopal

Algebra Level 2

Solve for $x$ when $2^{x+1} + 2^{x-1} = 640.$

The answer is 8.

3 solutions

Chew-Seong Cheong
Aug 10, 2018

$\begin{aligned} 2^{x+1} + 2^{x-1} & = 640 \\ 2^{x-1}\left(2^2 + 1\right) & = 640 \\ 5 \times 2^{x-1} & = 640 \\ 2^{x-1} & = 128 = 2^7 \\ x-1 & = 7 \\ \implies x & = \boxed 8 \end{aligned}$

@Calvin Lin , When I entered 8 as the answer it first said it was incorrect before the page refreshed (almost instantaneously) and deemed that I was correct.

Mohammad Farhat - 2 years, 8 months ago

I deduced what the answer is just by being super intimate with my powers of two and knowing the limited options, but I actually really like this solution of being able to squeeze out a 2^2 by applying a little bit of factoring. That's actually a really clever way to squeeze out a finite value and reduce the number of terms with x in them.

Brian Bohan - 2 years, 7 months ago

Ram Mohith
Aug 8, 2018

$2^x \cdot 2 + \dfrac{2^x}{2} = 640$

$2^x \cdot 4 + 2^x = 1280$

$2^x \times (4 + 1) = 1280 \implies 2^x = \dfrac{1280}{5}$

$\implies 2^x = 256 = 2^8$

$\therefore x = 8$

Srinivasa Gopal
Aug 8, 2018

Let 2^(X-1) = T. 2^(X+1) = 2^2* T = 4T. So the equation 2^(X+1) + 2^(X-1) = 640 reduces to 5T = 640 therefore T = 128.

So 2^(X-1) = 128 or in other words X- 1 = 7 , hence X = 8.

An algebra problem by Srinivasa Gopal

The answer is 8.

3 solutions

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