It's easier to start by squaring both sides and reducing the expression that was under the square root. We get the equation: {(2^x-9^x)*(2^x-9^x)}/{9^x*(2^x-9^x)}=1 >> We then simplify the left-hand side: (2^x-9^x)/9^x=1 Simplifying further, we get (2/9)^x-1=1 so (2/9)^x=2 Now we take the logarithm of both sides and get x*log(2/9)=log(2) >> x=log(2)/{log(2)-log(9)}. Changing the base of the logarithm makes no sense, because in this way we can use the logarithm with any base.
It's easier to start by squaring both sides and reducing the expression that was under the square root. We get the equation: {(2^x-9^x)*(2^x-9^x)}/{9^x*(2^x-9^x)}=1 >> We then simplify the left-hand side:
(2^x-9^x)/9^x=1 Simplifying further, we get (2/9)^x-1=1 so (2/9)^x=2 Now we take the logarithm of both sides and get x*log(2/9)=log(2) >> x=log(2)/{log(2)-log(9)}. Changing the base of the logarithm makes no sense, because in this way we can use the logarithm with any base.
(2^x-9^x)/9^x=1 , 2^x-9^x=9^x , 2^x=2*9^x , (2/9)^x=2 , x=log2/log(2/9) ,
Sorry, but x=0 result to 0 in denominator.
Log 9 = 2 log 3