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

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

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

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did you figure this out? if not i can send you an answer.
no and that would be great!
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.
\[5^x+\frac{1}{5}^{1-x}=\frac{1}{25^x}\]
add the terms on the left: \[\frac{5^{x+1-x}+1}{5^{1-x}}=\frac{5^1+1}{5^{1-x}}=\frac{6}{5^{1-x}}\]\]
Go ahead satellite :D
i am a slow typer.
so we have \[\frac{6}{x^{1-x}}=\frac{1}{25^x}\]
the right hand side is \[\frac{1}{5^{2x}}\]
\[\frac{6}{5^{1-x}}= \frac{1}{5^{2x}}\] \[6\times 5^{2x}=5^{1-x}\]
take the log of both sides to get the variable out of the exponent: \[ln(6\times 5^{2x})=ln(5^{1-x})\] \[ln(6) + 2xln(5) = (1-x)ln(5)\]
now it is algebra from here on in remembering that ln(6) and ln(5) are constants.
\[ln(6) +2xln(5)=ln(5)-xln(5)\] \[2xln(5)+xln(5)=ln(5)-ln(6)\] \[3ln(5)x=ln(5)-ln(x)\] \[x=\frac{ln(5)-ln(6)}{3ln(5)}\]
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 \(125^x=\frac{5}{6}\) \(x=\ln(5/6)/\ln(125)\)
if you choose you can rewrite this as \[\frac{ln(\frac{5}{6})}{3ln(5)}\]
watchman much snappier as usual. how it is going?
Awesome! Thank you!
Hrm. I woulda just: \[5^x + \frac{1}{5^{1-x}} = \frac{1}{25^x}\]\[\implies 5^{x+1} + 5^x = \frac{1}{5^x}\] \[\implies 5^x(5+1) = \frac{1}{5^x} \implies 5^{2x} = \frac{1}{6}\]\[ \implies x(ln\ 25) = (ln\ 1) - (ln\ 6)\]\[\implies x = \frac{(ln\ 1) - (ln\ 6)}{(ln\ 25)}\]
let me know if you do not understand any step
Not sure if that's the same as what you did.
It's not. Hrm.. now to see what went wrong.
must not be the same because we got different answers. fairly certain of mine
The \(5^{x+1}\) should be \(5^{x-1}\) Polpak.
\[\frac{1}{5}^{1-x}\neq 5^{x+1}\]
I multiplied the whole thing by 5. \[5^x + 5^{-1+x} = 5^{-2x}\] \[\implies 5^{x+1} + 5^x = 5^{-2x + 1}\]
It was the right side that I screwed up.
I canceled half the 5's from the \(1/25^x \) when I should have only cancelled one of them.
So.. \[5^{x+1} + 5^x = 5^{1-2x}\]\[\implies 5^x(5+1) = 5^{1-2x}\]\[\implies x(ln\ 5) + (ln\ 6) = (1-2x)(ln\ 5)\]\[\implies x = \frac{(ln\ 5) - (ln\ 6)}{3(ln\ 5)}\]
yeah, that's better.

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