Question

solve5^x+1/5^1-x=1/25^x

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

no and that would be great!

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

Go ahead satellite :D

Answer 10

i am a slow typer.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

\(125^x=\frac{5}{6}\)
\(x=\ln(5/6)/\ln(125)\)

Answer 18

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

Answer 19

watchman much snappier as usual. how it is going?

Answer 20

Awesome! Thank you!

Answer 21

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

Answer 22

let me know if you do not understand any step

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

The \(5^{x+1}\) should be \(5^{x-1}\) Polpak.

Answer 27

\[\frac{1}{5}^{1-x}\neq 5^{x+1}\]

Answer 28

I multiplied the whole thing by 5.

\[5^x + 5^{-1+x} = 5^{-2x}\]
\[\implies 5^{x+1} + 5^x = 5^{-2x + 1}\]

Answer 29

It was the right side that I screwed up.

Answer 30

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

Answer 31

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

Answer 32

yeah, that's better.