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Mathematics 10 Online
OpenStudy (anonymous):

Find the solution.

OpenStudy (anonymous):

\[e ^{-2x}=7\] \[\ln e ^{-2x}=\ln 7\] ?

OpenStudy (anonymous):

Please show step by step?

OpenStudy (anonymous):

\[ln[ e^x] = x\]

OpenStudy (anonymous):

\[ e^{-2x}=7\] \[\ln{[e^{-2x}]}=\ln{7}\] \[-2x =\ln{7}\] \[x = \frac{\ln{7}}{-2}\]

OpenStudy (anonymous):

where did the e go? @freewilly922

OpenStudy (anonymous):

The natural log (ln) is the inverse function of the exponential function e. \[e^{\ln{x}}=x\] and \[\ln{(e^x)} = x\]

OpenStudy (anonymous):

so anytime i have a function w/e, if i take the log or natural log, the e cancels?

OpenStudy (anonymous):

Yes, thats the whole reason for the natural log. So we can play with the exponents.

OpenStudy (anonymous):

Not the log thought. Natural log. Log usually implies log base 10. \[log_{10} 10^x = x\]

OpenStudy (anonymous):

gotcha! thanks :)

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