can someone help me.........

G

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.........i dont know how to do this problem

Solve for x: e^2x + 9^e-2x = 6


x =
 
AmandaLynn04 said:
.........i dont know how to do this problem

Solve for x: e^2x + 9^e-2x = 6


x =
Unable to translate.
Unable to believe this is a "Beginning Algebra" question.
 
Hello, AmandaLynn04!

I think you have a typo in your equation . . . I'll correct it.

Solve for \(\displaystyle x:\;\;e^{2x}\,+\,9e^{-2x}\:=\:6\)
Multiply by \(\displaystyle e^{2x}:\;\;e^{4x}\,+\,9\:=\:6e^{2x}\)

Then we have: .\(\displaystyle e^{4x}\,-\,6e^{2x}\,+\,9\:=\:0\)

. . which factors: .\(\displaystyle (e^{2x}\,-\,3)^2\:=\:0\)

Then: .\(\displaystyle e^{2x}\,-\,3\:=\:0\;\;\Rightarrow\;\;e^{2x}\,=\,3\)

Take logs: .\(\displaystyle \ln(e^{2x})\:=\:\ln(3)\;\;\Rightarrow\;\;2x\cdot\ln(e)\,=\,\ln(3)\;\;\Rightarrow\;\;2x\cdot1\,=\,\ln(3)\)

Therefore: .\(\displaystyle 2x\,=\,\ln(3)\;\;\Rightarrow\;\;x\,=\,\frac{\ln(3)}{2}\)
 
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