welcome back math prob: 2^(X+1) + 2^(X - 1) = 320

~*alphAOmega*~

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Sep 2, 2006
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lol just got back to school, beginning to get the math cogs into gear again :wink: so was just scrolling through some problems and got completly dumbfounded doing this sum :-

2^(X+1) + 2^(X - 1) = 320

what i did was i used the laws of exponents to make 2^(x + 1) into 2^X * 2
and 2^(X - 1) into 2^X * 1/2 ( as 2^-1 = 1/2^1)

so it comes out like

(2^X * 2) + (2^X * 1/2) = 320

now im stuck... :?
 
\(\displaystyle \L 2^{x+1} + 2^{x-1} = 320\)

\(\displaystyle \L 2^{x+2} + 2^x = 640\)

\(\displaystyle \L 2^x*2^2 + 2^x = 640\)

\(\displaystyle \L 2^x(2^2 + 1) = 640\)

\(\displaystyle \L 2^x = \frac{640}{5}\)

\(\displaystyle \L 2^x = 128\)

\(\displaystyle \L x = 7\)
 
Hint: What did he do to the equation to go from "=320" to "=640"?

Eliz.
 
~*alphAOmega*~ said:
\(\displaystyle \L 2^x*2^2 + 2^x = 640\)

how did you get that..? from the earlier step..? :roll:

2a+b=2a2b\displaystyle 2^{a+b} = 2^a*2^b

double :roll:
 
yup yup :) was just solving it and got it :wink:

he did 320 / .5

:)

thanks, lol i was working at this problem for 40 whole miniutes :(
 
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