Solve the equation $\displaystyle 3^{x+3}-3^{x+2}=2$
I seem to be having trouble working out this one... can someone show me what I did wrong?
$\displaystyle 3^{x+3}-3^{x+2}=2$
$\displaystyle \log{3^{x+3}}-\log{3^{x+2}}=\log{2}$
$\displaystyle (x+3)\log{3}-(x+2)\log{3}=\log{2}$
$\displaystyle x\log{3} + 3\log{3} - x\log{3} - 2\log{3} = \log{2}$
$\displaystyle x\log{3}- x\log{3} = \log{2}- 3\log{3} + 2\log{3}$
The x's cancel eachother out... and I don't know where went wrong.
(By the way, x is a whole number)
Another way to do it: by the laws of exponents, $\displaystyle 3^{x+ 3}= (3^x)(3^3)= 27(3^x)$ and $\displaystyle 3^{x+ 2}= (3^x)(3^2)= 9(3^x)$. The equation becomes $\displaystyle 3^{x+3}- 3^{x+ 2}= 27(3^x)- 9(3^x)= 18 (3^x)= 2$ so that $\displaystyle 3^x= \frac{2}{18}= \frac{1}{9}$. Since $\displaystyle 9= 3^2$, $\displaystyle \frac{1}{9}= 3^{-2}$ so we have, finally, $\displaystyle 3^x= 3^{-2}$ and x= -2.