Ask your own question, for FREE!
Physics 18 Online
OpenStudy (babyslapmafro):

Please help me find x for the following (click to see equation).

OpenStudy (babyslapmafro):

\[2e^x-1=0\]

OpenStudy (babyslapmafro):

\[2e^x=1\]

OpenStudy (babyslapmafro):

\[e^x=1/2\]

OpenStudy (babyslapmafro):

Now, how do I find x?

OpenStudy (anonymous):

take the natural log of both sides. ln(e^x) is just x

OpenStudy (babyslapmafro):

\[\ln(e^x)=\ln(1/2)\]

OpenStudy (babyslapmafro):

\[x=\ln(1/2)\]

OpenStudy (anonymous):

yes

OpenStudy (babyslapmafro):

the correct answer in the book is -ln2

OpenStudy (babyslapmafro):

i understand it is the same value...

OpenStudy (babyslapmafro):

But I am using the x value is a critical value to find the relative extrema of a function, does it matter if I use ln(1/2) or -ln2?

OpenStudy (babyslapmafro):

as* a critical value

OpenStudy (babyslapmafro):

@Algebraic!

OpenStudy (anonymous):

yeah, same thing: ln(1/2) = ln (2^-1) = - ln(2) all different ways of expressing the same thing. check with your calculator if you like..

OpenStudy (anonymous):

"But I am using the x value is a critical value to find the relative extrema of a function, does it matter if I use ln(1/2) or -ln2?" no, they are all exactly the same number.

OpenStudy (babyslapmafro):

ok thanks

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!