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.........i dont know how to do this problem
Solve for x: e^2x + 9^e-2x = 6
x =
Solve for x: e^2x + 9^e-2x = 6
x =
Hello, AmandaLynn04!
I think you have a typo in your equation . . . I'll correct it.
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: .\(\displaystyle 2x\,=\,\ln(3)\;\;\Rightarrow\;\;x\,=\,\frac{\ln(3)}{2}\)
I think you have a typo in your equation . . . I'll correct it.
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: .\(\displaystyle 2x\,=\,\ln(3)\;\;\Rightarrow\;\;x\,=\,\frac{\ln(3)}{2}\)