# Math Help - Please solve the equation for "Y"?

1. ## Please solve the equation for "Y"?

Hello,

Most of these I have already solved, and have included the answer I got! Please tell me if I have gotten the answer right. Thank you. Some of these I do need a bit of help with.
If you don't have time to do all of them and check them, you can just do a couple! Anything will help! Thank you.

So let's start.

1. 3y^e = x-5
I got y = e root((x-5)/3)

2. lny^3 = 3x
I got y = 3rd root(e^3x)

3. 3 + ln(abs y) = -x
I got e^(-x-3) = y

4. ln(y-3) = x-2

5. 3e^y = x-5
I got y = ln(x-5)/3

Thank you!

2. The first 3 seem correct to me.

4. ln(y-3) = x-2

$\ln{(y-3)} = x-2$

$e^{\ln{(y-3)}} = e^{x-2}$

$y-3 = e^{x-2}$

$y = e^{x-2} + 3$

5 seems incorrect. Here's what got:

$3e^y = x-5$

$e^y = \frac{x-5}{3}$

$\ln(e^y) = \ln(\frac{x-5}{3})$

$y= \ln(\frac{x-5}{3})$

3. 1) seems okay.

2) y= e^(3rd root ( 3x) )

3) y = +- e^(-x-3)

4) y = e^(x-2) + 3

5) y = ln((x-5)/3) which I think you meant anyway.

EDIT : I see now that by lny^3 you meant ln(y^3), in which case your answer is correct.

4. Wow. Thank you so much. I made one mistake, for number 4, the actual question was ln(abs(y-3)) = x-2 . So would the answer be plus/minus? Thank you!

5. Actually the second one can be simplified down to just:

y = e^x