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Q. Prove


My try: expand and got:


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

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$5^{x^2-1}$ is not correct, you should have there $5^{(x+1)+(x-1)}=5^{2x}$. \\ Hint for the original problem: Try to factor out $5^{2x}$ or $5^{2x-2}$ or something similar in both numerator and denominator. – Martin Sleziak Jan 25 '12 at 9:53
@MartinSleziak, ah I am always confused about the +-*/ stuff ... – Jiew Meng Jan 25 '12 at 10:13
up vote 4 down vote accepted

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

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OH nice, but can you factorise $(u/5+5u)^2 = u^2(1/5+5)^2$? – Jiew Meng Jan 25 '12 at 10:41
Oops, you were faster than me at writing your own idea! – Gigili Jan 25 '12 at 10:42
@jiewmeng, $(u/5+5u)^2 = (u(1/5+5))^2 = u^2(1/5+5)^2$. – lhf Jan 25 '12 at 10:48
I also tried substituting $u = 5^={2x}$ just now but got a wrong answer, why is that? – Jiew Meng Jan 25 '12 at 11:05
@jiewmeng, the denominator in the last line but one should be $u+5^4u$. – lhf Jan 25 '12 at 11:11

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.

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How did you the the "non-x" (right most, no x variables) part? – Jiew Meng Jan 25 '12 at 10:08
I am not sure I understand what you mean. Do you mean why ${(5^{x-1}+5^{x+1})^2}={(5^{x-1})^2(1+5^2)^2}$? . This simply follows from $5^{x+1}=(5^2)5^{x-1}$. You can reason similarly for the denominator. – Henrik Jan 25 '12 at 10:23
Then $(5^2)5x−1 = (1+5^2)^2$ how did you get this? Sorry, I am abit blur here – Jiew Meng Jan 25 '12 at 10:31
everything i squared in the numerator. remember that $(ab+ac)^2=a^2(b+c)^2$. – Henrik Jan 25 '12 at 10:39

To find $x$ from



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


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

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

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