I have quick question. e^(x-1) = 25 I try to set it up as: x-1 ln = ln25 Where to from there? :?
W WTF? Junior Member Joined Sep 16, 2005 Messages 95 Mar 7, 2006 #1 I have quick question. e^(x-1) = 25 I try to set it up as: x-1 ln = ln25 Where to from there? :?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 7, 2006 #3 Hello, WTF? You're almost there . . . \(\displaystyle e^{x-1}\; =\; 25\) I try to set it up as: \(\displaystyle \,(x-1)\ln(e)\:=\:\ln(25)\) Where to from there? \(\displaystyle \;\) . . . solve for x Click to expand... It may be clearer if you include all the steps . . . We have: \(\displaystyle \,e^{^{x-1}}\:=\:25\) Take logs: \(\displaystyle \,\ln(e^{^{x-1}})\:=\;\ln(25)\) Then: \(\displaystyle \,(x-1)\cdot\ln(e)\:=\:\ln(25)\) Since \(\displaystyle \ln(e)\,=\,1\), we have: \(\displaystyle \,x\,-\,1\:=\:\ln(25)\;\;\Rightarrow\;\;x\:=\:\ln(25)\,+\,1\)
Hello, WTF? You're almost there . . . \(\displaystyle e^{x-1}\; =\; 25\) I try to set it up as: \(\displaystyle \,(x-1)\ln(e)\:=\:\ln(25)\) Where to from there? \(\displaystyle \;\) . . . solve for x Click to expand... It may be clearer if you include all the steps . . . We have: \(\displaystyle \,e^{^{x-1}}\:=\:25\) Take logs: \(\displaystyle \,\ln(e^{^{x-1}})\:=\;\ln(25)\) Then: \(\displaystyle \,(x-1)\cdot\ln(e)\:=\:\ln(25)\) Since \(\displaystyle \ln(e)\,=\,1\), we have: \(\displaystyle \,x\,-\,1\:=\:\ln(25)\;\;\Rightarrow\;\;x\:=\:\ln(25)\,+\,1\)