# Thread: Problem with solving equation

1. ## Problem with solving equation

Having trouble with a section in my summer assaignment, and my teacher doesnt respond to his e-mails... If someone could help with these few questions, it would help a lot. Thanks.

Solve for y in terms of t:

-2e^(-y/3)=t+4

-t/4)-ln3=ln(20-y)

I know for the 2nd question you need to combine to take the anti-log(I beleive), but I just dont know where. And I have no idea where to start on the first problem.

2. ## Re: Problem with solving equation

For the first one, take the $\displaystyle \ln$ of both sides so you get:
$\displaystyle \ln\left(2e^{-\frac{-y}{3}}\right)=\ln(t+4)$
Now use the fact $\displaystyle \log(a^b)=b\cdot \log(a)$

3. ## Re: Problem with solving equation

You put a negative symbol before the (y/3). Could you tell me where you got that from?
Did you get it from the -2e from the begining and moved it to the exponent when you multiplied by the ln?

Also that would make -y/3 = (ln2e) X ((ln t+4))? e would have to be raised by somethig though, so I know that isnt right... but I'm hoping its something along those lines.

4. ## Re: Problem with solving equation

Sorry, I didn't see that. That's indeed a mistake. Indeed you get:
$\displaystyle \ln(2)-\frac{y}{3}=\ln(t+4) \Leftrightarrow -\frac{y}{3}=\ln(t+4)-\ln(2) \Leftrightarrow -\frac{y}{3}=\ln\left(\frac{t+4}{2}\right) \Leftrightarrow y=-3\ln\left(\frac{t+4}{2}\right)$

For the second one I would use the definition:
$\displaystyle \log_a(x)=y \Leftrightarrow a^y=x$
So in this case:
$\displaystyle \frac{t}{4}-\ln(3)=\ln(20-y) \Leftrightarrow e^{\frac{t}{4}-\ln(3)}=20-y \Leftrightarrow ...$

5. ## Re: Problem with solving equation

Ahh, that makes sense. Thank you so much.

6. ## Re: Problem with solving equation

You're welcome!
Notice you can simplify the second one:
$\displaystyle \frac{t}{4}-\ln(3)=\ln(20-y)$
$\displaystyle \Leftrightarrow e^{\frac{t}{4}-\ln(3)}=20-y$
$\displaystyle \Leftrightarrow y=20-e^{\frac{t}{4}-\ln(3}}$
$\displaystyle \Leftrightarrow y=20-\frac{e^{\frac{t}{4}}}{e^{\ln(3)}}$
$\displaystyle \Leftrightarrow y=20-\frac{e^{\frac{t}{4}}}{3}$
$\displaystyle \Leftrightarrow y=20-\frac{\sqrt[4]{t}}{3}$

7. ## Re: Problem with solving equation

That helps a lot! Thank you once again!