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~ :)
2^x+1 =9
how do i solve for this? thanks!
3 answers
sorry, the plus one is part of the exponent
2^(x+1) =9
is it possible to move the 1 to the otherside still?
2^(x+1) =9
is it possible to move the 1 to the otherside still?
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~ :)
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~ :)
8 answers