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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