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## anonymous 5 years ago solve5^x+1/5^1-x=1/25^x

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1. anonymous

First combine your like terms. What do you have after that?

2. anonymous

5^x+1/5^1-x = 5^x+1-1+x = 5^2x 1/25^x = 1/5^2x 5^2x = 1/5^2x 5^2x times 5^2x = 1 5^4x = 5^0 4x = 0 x=0

3. anonymous

I don't think so cause if I plug in 0 to the original equation I get: $5^0 + 1/5^1 - 0 = 1/25^0$$\implies 1 + 1/5 = 1$$\implies 1 = 4/5$$\implies 5 = 4$ So yeah.

4. anonymous

did you figure this out? if not i can send you an answer.

5. anonymous

no and that would be great!

6. anonymous

ok let me go slow because it take a while for me to type this in. i will do it step by step. there may be a snap way to do this but i don't see it.

7. anonymous

$5^x+\frac{1}{5}^{1-x}=\frac{1}{25^x}$

8. anonymous

add the terms on the left: $\frac{5^{x+1-x}+1}{5^{1-x}}=\frac{5^1+1}{5^{1-x}}=\frac{6}{5^{1-x}}$\]

9. watchmath

Go ahead satellite :D

10. anonymous

i am a slow typer.

11. anonymous

so we have $\frac{6}{x^{1-x}}=\frac{1}{25^x}$

12. anonymous

the right hand side is $\frac{1}{5^{2x}}$

13. anonymous

$\frac{6}{5^{1-x}}= \frac{1}{5^{2x}}$ $6\times 5^{2x}=5^{1-x}$

14. anonymous

take the log of both sides to get the variable out of the exponent: $ln(6\times 5^{2x})=ln(5^{1-x})$ $ln(6) + 2xln(5) = (1-x)ln(5)$

15. anonymous

now it is algebra from here on in remembering that ln(6) and ln(5) are constants.

16. anonymous

$ln(6) +2xln(5)=ln(5)-xln(5)$ $2xln(5)+xln(5)=ln(5)-ln(6)$ $3ln(5)x=ln(5)-ln(x)$ $x=\frac{ln(5)-ln(6)}{3ln(5)}$

17. watchmath

Agree! :) $$5^{x}+5^{x-1}=5^{-2x}$$ $$(1+\frac{1}{5})5^x=5^{-2x}$$ $$\frac{6}{5}5^x=5^{-2x}$$ Multiply by $$5^{2x}$$ we have $$\frac{6}{5}5^{3x}=1$$ Then $$125^x=\frac{5}{6}$$ $$x=\ln(5/6)/\ln(125)$$

18. anonymous

if you choose you can rewrite this as $\frac{ln(\frac{5}{6})}{3ln(5)}$

19. anonymous

watchman much snappier as usual. how it is going?

20. anonymous

Awesome! Thank you!

21. anonymous

Hrm. I woulda just: $5^x + \frac{1}{5^{1-x}} = \frac{1}{25^x}$$\implies 5^{x+1} + 5^x = \frac{1}{5^x}$ $\implies 5^x(5+1) = \frac{1}{5^x} \implies 5^{2x} = \frac{1}{6}$$\implies x(ln\ 25) = (ln\ 1) - (ln\ 6)$$\implies x = \frac{(ln\ 1) - (ln\ 6)}{(ln\ 25)}$

22. anonymous

let me know if you do not understand any step

23. anonymous

Not sure if that's the same as what you did.

24. anonymous

It's not. Hrm.. now to see what went wrong.

25. anonymous

must not be the same because we got different answers. fairly certain of mine

26. watchmath

The $$5^{x+1}$$ should be $$5^{x-1}$$ Polpak.

27. anonymous

$\frac{1}{5}^{1-x}\neq 5^{x+1}$

28. anonymous

I multiplied the whole thing by 5. $5^x + 5^{-1+x} = 5^{-2x}$ $\implies 5^{x+1} + 5^x = 5^{-2x + 1}$

29. anonymous

It was the right side that I screwed up.

30. anonymous

I canceled half the 5's from the $$1/25^x$$ when I should have only cancelled one of them.

31. anonymous

So.. $5^{x+1} + 5^x = 5^{1-2x}$$\implies 5^x(5+1) = 5^{1-2x}$$\implies x(ln\ 5) + (ln\ 6) = (1-2x)(ln\ 5)$$\implies x = \frac{(ln\ 5) - (ln\ 6)}{3(ln\ 5)}$

32. anonymous

yeah, that's better.

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