Posted by **greeny** on Monday, May 21, 2012 at 10:35pm.

2^x+1 =9

how do i solve for this? thanks!

- math -
**Jai**, Monday, May 21, 2012 at 10:38pm
2^x + 1 = 9

2^x = 9 - 1

2^x = 8

then we rewrite 8 as power of 2:

2^x = 2^3

finally, since they have the same base (which is 2) we equate their exponents:

x = 3

hope this helps~ :)

- math -
**greeny**, Monday, May 21, 2012 at 10:44pm
sorry, the plus one is part of the exponent

2^(x+1) =9

is it possible to move the 1 to the otherside still?

- math -
**Jai**, Monday, May 21, 2012 at 10:52pm
Oh sorry. I thought it's outside. Nope, it can't be moved outside.

2^(x+1) = 9

2^(x+1) = 3^2

then we take the log of both sides:

(x+1)*(log 2) = 2*(log 3)

x + 1 = 2(log 3) / (log 2)

x = [2(log 3) / (log 2)] - 1

or we can also rewrite the 1 as (log 2)/(log 2), so

x = [2(log 3) - (log 2)] / (log 2)

hope this helps~ :)

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