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Mathematics 12 Online
OpenStudy (anonymous):

e^3x-5 e^2x=135

OpenStudy (anonymous):

e^5x-5=135 take log on both sides

OpenStudy (anonymous):

\[\log e^{5x-5}=\log 135\] can you do it?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

cool

OpenStudy (anonymous):

no

OpenStudy (anonymous):

yes wait a sec

OpenStudy (anonymous):

the answer in the back of the book is 1.9811 and i cant seem to get it

OpenStudy (anonymous):

\[e^{3x-5}*e^{2x}=e^{3x+2x-5}=e^{5x-5}\]

OpenStudy (anonymous):

|dw:1351565533449:dw|

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