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Mathematics 24 Online
OpenStudy (anonymous):

ln x = -4

OpenStudy (anonymous):

Hello Jenn... x = e^-4 Solve it to get the value for x

OpenStudy (anonymous):

so the formula that you are using is what? Please remind me!

OpenStudy (anonymous):

He's multiplying both sides by e

jimthompson5910 (jim_thompson5910):

ln(x)=y becomes e^y = x

jimthompson5910 (jim_thompson5910):

not quite outkast

OpenStudy (anonymous):

ln x => log_e x log to base e ln x = b then x = e^b

OpenStudy (anonymous):

so the numeric answer is: .018315638 ? Yes?

OpenStudy (anonymous):

not quite are you teling me that e^ln(x) =e^-4

OpenStudy (anonymous):

Exacly!!

jimthompson5910 (jim_thompson5910):

yes, that's more like it

jimthompson5910 (jim_thompson5910):

and e^ln(x) = x

OpenStudy (anonymous):

oh i wrote it differently but i mean't what is above

jimthompson5910 (jim_thompson5910):

oh ok

OpenStudy (anonymous):

Joe shouldn't you be studying for your tutoring job haha

OpenStudy (anonymous):

wow! such a debate! I hope I have it now! THANKS ALL!

OpenStudy (anonymous):

\[ln(x) = -4 \]\[\implies log_e(x) = -4\]\[ \implies e^{-4} = x\]\[\implies x = \frac{1}{e^4} \approx .018316\]

OpenStudy (anonymous):

I hate the log

OpenStudy (anonymous):

=]

OpenStudy (anonymous):

Everything you need to know about logs you learned in exponentiation. Oh, and this \[log_b(a) = k \iff b^k = a\]

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