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OpenStudy (anonymous):
ln x = -4
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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
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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
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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!
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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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