Why is $3^{(x-5)} + 3^{(x-7)} + 3^{(x-9)} = 91$?

Question

So far I think that this is somehow related to that $(x-7) - (x-5) = (x-9) - (x-7) = 2$, but is it ?
What steps do you take to add $3^{x-5} + 3^{x-7} + 3^{x-9}$ up ?
Thank you!

Answer 1

Notice that: \begin{align*} 3^{x-5} + 3^{x-7} + 3^{x-9} &= 3^{4 + (x-9)} + 3^{2 + (x-9)} + 3^{x-9} \\ &= (3^4)3^{x-9} + (3^2)3^{x-9} + 3^{x-9} \\ &= (81)3^{x-9} + (9)3^{x-9} + (1)3^{x-9} \\ &= (81 + 9 + 1)3^{x-9} \\ &= (91)3^{x-9} \\ \end{align*}

Answer 2

This is false.

Take $x=0$, then $3^{x-5} + 3^{x-7} + 3^{x-9}=\frac{1}{3^5}+\frac{1}{3^7}+\frac{1}{3^9}\neq91$

Answer 3

It will be $91$ only if $x=9$, you have not specified what $x$ is, or do we have to find $x$?

In that case $3^{(x-5)} + 3^{(x-7)} + 3^{(x-9)} = 3^{(x-9)} (1+3^2+3^4)=3^{(x-9)}(91)=91 \implies x=9 $