Melissa_Something
  • Melissa_Something
I got the answer, I just dont know how. LOG help :(
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
here to help m8
Melissa_Something
  • Melissa_Something
\[e^x-5=(1/e^4)^x+2\]
Melissa_Something
  • Melissa_Something
the x+2 is in the exponent with the x :/ Lol thanks

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More answers

anonymous
  • anonymous
u should see if tiger algebra will give the answer
Melissa_Something
  • Melissa_Something
My teacher said something about how a power and power cancel, and her next step was this \[e=e^-4(x+2)\]
anonymous
  • anonymous
just sayin retipe it here if u can and i will solve
anonymous
  • anonymous
retype
Melissa_Something
  • Melissa_Something
Whats tiger algebra?
anonymous
  • anonymous
this is the website tiger-algebra.com
anonymous
  • anonymous
type in your browser
Melissa_Something
  • Melissa_Something
I have the answer, just dont know how
anonymous
  • anonymous
and post ypur answer
anonymous
  • anonymous
your
anonymous
  • anonymous
didi u do it
anonymous
  • anonymous
that website will show u the steps too m8
Melissa_Something
  • Melissa_Something
It cant do what I needed :/
anonymous
  • anonymous
damn what do u need m8
anonymous
  • anonymous
sorry im part russian german and african american so im good in some things most i cant really do
Nnesha
  • Nnesha
is this ur question \[\huge\rm e^{x-5}=(\frac{1}{e^4})^{x+2}\] @Melissa_Something
Nnesha
  • Nnesha
i don't know if the left side is e^{x-5} or e^(x) - 5
Nnesha
  • Nnesha
but right side you can move e^4 to the numerator just like we did on the previous post
Nnesha
  • Nnesha
is this ur question \[\huge\rm e^{x-5}=(\frac{1}{e^4})^{x+2}\] `OR` \[\huge\rm e^x-5 =(\frac{1}{e^4})^{(x+2)}\]
Nnesha
  • Nnesha
here is an example \[\huge\rm \frac{1}{x^{-m}}= x^{m}\] when we flip the fraction sign of the exponent would change so \[\frac{1}{e^{4}}=?\]
Nnesha
  • Nnesha
and yes `e` and `ln` cancel each other out \[\large\rm e^{\ln x} = x\] \[\large\rm \cancel{e^{\ln} x} = x\]

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