# Please solve the equation for "Y"? (a few more problems!)

• Sep 18th 2010, 11:24 PM
Please solve the equation for "Y"? (a few more problems!)

Hello,

I wasn't completly sure on the ones I have solved.

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.

1. (e^-y) + (e^y) = 2

2. (x^2)(ln y) - xlny = 3
I got y = e^(3/(x^2)-x)

3. (ln y )^3 = 5^x
I got e^third root(5x)

4. ln(1+ln y ) = 3
I WASN'T SURE ON THIS ONE. PLEASE HELP! (I didn't know if I was supposed to distribute the "ln" or what. Thank you.

5. (e^(-2y)) + 1 = 3e^x
I got y = (ln3(e^x) - 1)/(-2lne)

6. y^x = lnx + 2
I got y = x root (lnx + 2)

Thank you everyone!
• Sep 18th 2010, 11:29 PM
Prove It
$e^{-y} + e^y = 2$

$\frac{1}{e^y} + e^y = 2$

$1 + e^{2y} = 2e^y$ after multiplying both sides by $e^{y}$

$e^{2y} - 2e^y + 1 = 0$

$Y^2 - 2Y + 1 = 0$ if we let $Y = e^y$

$(Y-1)^2 = 0$

$Y-1 = 0$

$Y = 1$

$e^y = 1$

$y = \ln{1}$

$y= 0$.
• Sep 18th 2010, 11:31 PM
Prove It
Quote:

Hello,

I wasn't completly sure on the ones I have solved.

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.

1. (e^-y) + (e^y) = 2

2. (x^2)(ln y) - xlny = 3
I got y = e^(3/(x^2)-x)

3. (ln y )^3 = 5^x
I got e^third root(5x)

4. ln(1+ln y ) = 3
I WASN'T SURE ON THIS ONE. PLEASE HELP! (I didn't know if I was supposed to distribute the "ln" or what. Thank you.

5. (e^(-2y)) + 1 = 3e^x
I got y = (ln3(e^x) - 1)/(-2lne)

6. y^x = lnx + 2
I got y = x root (lnx + 2)

Thank you everyone!

$\ln{(1 + \ln{y})} = 3$

$1 + \ln{y} = e^3$

$\ln{y} = e^3 - 1$

$y = e^{e^3 - 1}$.
• Sep 18th 2010, 11:35 PM
Educated
The rest (2,3,5 and 6) seem correct to me.

For number 5:

y = (ln3(e^x) - 1)/(-2lne) can be simplified to: y = (ln3(e^x) - 1)/(-2)

Remember ln(e^1) = 1
• Sep 18th 2010, 11:42 PM