anonymous
  • anonymous
solve5^x+1/5^1-x=1/25^x
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
First combine your like terms. What do you have after that?
anonymous
  • 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
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.