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