Prove $\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$

Question

Q. Prove

$$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{338}{313}$$

My try: expand and got:

$$\frac{5^{2x-2}+2(5^{x^2-1})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}$$

Now what? I find my pre-calculus skills esp with Indices, Logarithms & Trigo lacking ... need to know how to apply the formulas more

Answer 1

Let $u=5^x$. Then $u^2=25^x$ and $$ \frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{(u/5+5u)^2}{u^2/25+25u^2}=\frac{u^2(1/5+5)^2}{u^2(1/25+25)}=\frac{(1/5+5)^2}{(1/25+25)}=\frac{338}{313} $$

Answer 2

Factor as: $$\frac{(5^{x-1}+5^{x+1})^2}{25^{x-1}+25^{x+1}}=\frac{(5^{x-1})^2(1+5^2)^2}{(5^{x-1})^2(1+25^2)}$$

Then simplify.

Answer 3

To find $x$ from

$\displaystyle\frac{5^{2x-2}+2(5^{2x})+5^{2x+2}}{5^{2x-2}+5^{2x+2}}=\frac{338}{313}$

i.e.

$\displaystyle 1+\frac{2\times 5^{2x}}{5^{2x-2}+5^{2x+2}}=1+\frac{25}{313}$

i.e.

$\displaystyle\frac{2\times 5^{2x}}{5^{2x-2}+5^{2x+2}}=\frac{25}{313}$

i.e. $\displaystyle\frac{2}{5^{-2}+5^2}=\frac{25}{313}$, an identity. So the above equation is valid for all $x\epsilon \mathbb{R}$.