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Mathematics 6 Online
OpenStudy (anonymous):

how do you simplify ln square root e, and is the answer e^2?

OpenStudy (anonymous):

no it is \[\frac{1}{2}\]

OpenStudy (anonymous):

\[ln(x)=y\] same as \[e^y=x\] \[ln(\sqrt{e})=ln(e^{\frac{1}{2}})\]

OpenStudy (anonymous):

this is starting to make a lot more sense thanx a lot!

OpenStudy (anonymous):

and since clearly \[e^{y}=e^{\frac{1}{2}}\] makes \[y=\frac{1}{2}\] we get \[ln(e^{\frac{1}{2}})=\frac{1}{2}\]

OpenStudy (anonymous):

this of it this way. \[ln(x)\] is the power you raise e to to get x. therefore \[ln(e^y)=y\]

OpenStudy (anonymous):

just like \[log_3(3^2)=2\]

OpenStudy (anonymous):

and \[log_{10}(.001)=-3\] since \[10^{-3}=.001\]

OpenStudy (anonymous):

10 ^-4 = 0.0001?

OpenStudy (anonymous):

yes, since \[10^{-4}=\frac{1}{10^4}=\frac{1}{10000}=.0001\]

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