Solve leaving answers in terms ofeor ln.

a) e^(5x-3)=10

b) 5^(3+x) = 20^x-3

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- Jul 28th 2010, 06:49 PMdanielh9103Precalc HELP, ASAP. No idea, Please explain
Solve leaving answers in terms of

*e*or ln.

a) e^(5x-3)=10

b) 5^(3+x) = 20^x-3 - Jul 28th 2010, 06:56 PMlvleph
$\displaystyle \ln$ basically undoes $\displaystyle e$. So if you want to isolate the exponent of $\displaystyle e^x$ we can take the natural log like so $\displaystyle \ln e^x = x$. Now don't forget whatever you do to one side of an equation you have to do to the other side of the equation. Also, don't forget that $\displaystyle \ln a^x = x\cdot \ln a$. Make an attempt at these problems and post what you come up with.

- Jul 28th 2010, 07:03 PMdanielh9103
alright thank you!

is this correct:

5x - 3 = ln 10

5x = ln 10 + 3

x = [(ln 10) + 3] / 5 - Jul 28th 2010, 07:05 PMlvleph
Yep.

- Jul 28th 2010, 07:15 PMdanielh9103
thank you, and for the second one, im stuck here but i have

ln5 (3+x) = ln 20 (x-3)

3 ln5 + x ln5 = x ln20 - 3 ln20

3 ln5 + 3 ln20 = x(ln20 - ln5)