OpenStudy (anonymous):

solve the equation for x. e^7-4x=6

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):

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, $a^n=b\Rightarrow n\log(a)=\log(b)$

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.

OpenStudy (anonymous):

oh okay! well thank you very much John_ES I can solve from here!

OpenStudy (john_es):

;)