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OpenStudy (john_es):
You mean,
\[e^{7-4x}=6?\]
OpenStudy (anonymous):
Yes, that's correct.
OpenStudy (anonymous):
I'm not sure how to go about solving this problem.
OpenStudy (john_es):
Do you know about logarithms?
OpenStudy (john_es):
I give you a hint. One of the properties of logarithms is
\[\log(a^n)=n\log(a)\]
So you can employ that in an equation like this,
\[a^n=b\Rightarrow n\log(a)=\log(b)\]
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OpenStudy (anonymous):
Well, yeah I guess so. I think that I should find a common base so that i can cancel them out and set the exponents to each other but i'm not sure, I could be wrong.
OpenStudy (john_es):
It would be better to use natural logarithms, because
\[\ln(e)=1\]
OpenStudy (anonymous):
Yes, I see that log is the inverse or is comparable to log, so using the equation would you get (7-4x)log(6)?
OpenStudy (john_es):
Well,
\[\ln(e^{7-4x})=\ln(6)\Rightarrow 7-4x=\ln(6)\]
You missed an =.
OpenStudy (john_es):
Then solve for x.
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OpenStudy (anonymous):
oh okay! well thank you very much John_ES I can solve from here!