i have been trying to work this question but i end up with $\displaystyle 3^{x+3} - 3 ^{x+1}$
can someone please me what i am doing wrong.
$\displaystyle 9^{x+2} = 9^2 \cdot 9^x$. Similarly, $\displaystyle 3^{x+1} = 3 \cdot 3^x$ and $\displaystyle 3^{x-1} = 3^{-1} \cdot 3^x$
Which gives $\displaystyle \dfrac{81 \cdot 9^x - 9^x}{3 \cdot 3^x - \dfrac{1}{3} \cdot3^x} = 10(3 \cdot 3^x)$
Factor and use the fact that $\displaystyle 9 = 3^2$
$\displaystyle \dfrac{(81-1) \cdot 9^x}{3^x \left(3-\dfrac{1}{3}\right)} = \dfrac{80 \cdot 9^x}{\dfrac{8}{3} \cdot 3^x} = 30 \cdot 10^{x}$
Once you do some cancelling you get:
$\displaystyle \dfrac{9^x}{3^x} = 10^{x}$
You know that $\displaystyle 9^x = (3^2)^x = 3^{2x}$ so rewrite the equation as
$\displaystyle \dfrac{3^{2x}}{3^x} = 3^x$
$\displaystyle 3^x = 10^x$ is trivial and by inspection only $\displaystyle x=0$ would work but you can verify using logarithms
Hello, sigma1!
$\displaystyle \text{Simplify: }\:\dfrac{9^{x+2} - 9^x}{3^{x+1} - 3^{x-1}} $
$\displaystyle \text{Numerator: }\;9^{x+2} - 9^x \:=\:(3^2)^{x+2} - (3^2)^x \:=\:3^{2x+4} - 3^{2x}$
. . . . . . . . $\displaystyle =\;3^{2x}(3^4 - 1) \;=\;80\cdot3^{2x}$
$\displaystyle \text{Denominator: }\;3^{x+1} - 3^{x-1} \;=\;3^{x-1}(3^2-1) \;=\;8\cdot3^{x-1}$
$\displaystyle \text{The problem becomes: }\;\dfrac{80\cdot3^{2x}}{8\cdot3^{x-1}} \;=\;10\cdot3^{2x-(x-1)} \;=\;10\cdot3^{x+1}$
$\displaystyle \displaystyle\frac{9^{x+2}-9^x}{3^{x+1}-3^{x-1}}=\frac{9^x\left(9^2-1\right)}{3^{x-1}\left(3^2-1\right)}=\frac{3^{2x}\left(9^2-1\right)}{3^{x-1}\left(3^2-1\right)}$
I think you are getting your answer as follows....
$\displaystyle \displaystyle\frac{9^{2x}-9^x}{3^{x+1}-3^{x-1}}=\frac{3^{2x+4}-3^{2x}}{3^{x+1}-3^{x-1}}$
Now you may be subtracting x+1 from 2x+4 and x-1 from 2x which is incorrect.
You can only do it that way if you have factors.