1. ## Exponential problem

Guys, Just a quick one:

Can anyone help me with e^((x+2lnx)) which is part of a bigger problem I'm working on?

I think that it is e^x + x^2 but apparently this is wrong. I can't see how to move on as I though if you took the anti-log this is what you'd get.

D

2. Originally Posted by dojo
Guys, Just a quick one:

Can anyone help me with e^((x+2lnx)) which is part of a bigger problem I'm working on?

I think that it is e^x * x^2 but apparently this is wrong.<<<<< correct! See my correction
I can't see how to move on as I though if you took the anti-log this is what you'd get.

D
...

3. Does that make it xe^2x then?

4. Originally Posted by dojo
Does that make it xe^2x then?
$e^{x+2ln(x)} = e^x \cdot e^{2\ln(x)} = e^x \cdot x^2$

You can't simplify this term any more.

5. Thanks! This is part of another question which I stuck on.

ln (3+2y) - ln (3-2y) = (x + 2lnx)

Express y in terms of x in the form y = f(x)

I can remove the logs and end up with :

(3+2y)/(3-2y) = e^x * x^2

But I'm unsure where to go next. If I multiply out the LSH I get in a mess on the RHS.

Where should I go from here?

Thanks D

6. Originally Posted by dojo
Thanks! This is part of another question which I stuck on.

ln (3+2y) - ln (3-2y) = (x + 2lnx)

Express y in terms of x in the form y = f(x)

I can remove the logs and end up with :

(3+2y)/(3-2y) = e^x * x^2

But I'm unsure where to go next. If I multiply out the LSH I get in a mess on the RHS.

Where should I go from here?

Thanks D
e^x and x^2 can be treated as a normal fraction. Multiply both sides by (3-2y) and expand before isolating y

7. Originally Posted by dojo
Thanks! This is part of another question which I stuck on.

ln (3+2y) - ln (3-2y) = (x + 2lnx)

Express y in terms of x in the form y = f(x)

I can remove the logs and end up with :

(3+2y)/(3-2y) = e^x * x^2

But I'm unsure where to go next. If I multiply out the LSH I get in a mess on the RHS.

Where should I go from here?

Thanks D
$\dfrac{3+2y}{3-2y} = e^x \cdot x^2$

Multiply by the denominator, seperate the terms: All terms containing y at the RHS, all other at the LHS:

$3+2y = (e^x \cdot x^2)(3-2y)~\implies~3-3(e^x \cdot x^2)=-2y-2y(e^x \cdot x^2)$

Now factor out y at the RHS:

$3-3(e^x \cdot x^2)=y(-2-2(e^x \cdot x^2))$

Divide by the bracket of the RHS to get y. Simplify this term a little bit more.

8. OK. I think I might have it, could you check through these calculations to see if I'm right? Excuse if this looks a it messy...

ln (3+2y) - ln (3-2y) = (x + 2lnx)

ln ((3+2y)/(3-2y) = (x + 2lnx)

(3+2y)/(3-2y) = e^x * x^2

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

3+2y = 9ex^2x - 6e^xyx^2 - 6e^xyx^2 + 4e^xy^2x^2

Simplified...

3+2y = ex^2x (9 - 8y^2)

(3+2y)/(9 - 8y^2) = ex^2x

(3-2)/y = -72ex^2x

1/(-72ex^2x) = y

9. Sorry, Ignore that last post. Your help in the reply before didn't open on my Mac. I can't see how you would tidy up the LHS though...

10. Originally Posted by dojo
OK. I think I might have it, could you check through these calculations to see if I'm right? Excuse if this looks a it messy...

ln (3+2y) - ln (3-2y) = (x + 2lnx)

ln ((3+2y)/(3-2y) = (x + 2lnx)

(3+2y)/(3-2y) = e^x * x^2

3+2y = (e^x * x^2) (3 - 2y) <<<<<< OK

3+2y = 9ex^2x - 6e^xyx^2 - 6e^xyx^2 + 4e^xy^2x^2 <<<<<<<< Sorry, but NO!

Simplified...

...
I'll take your last correct line:

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

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

3 - 3(e^x * x^2) = -2y- 2y(e^x * x^2)

3 - 3(e^x * x^2) = y(-2- 2(e^x * x^2))

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

$-\dfrac32 \cdot \dfrac{1 - (e^x * x^2)}{1+(e^x * x^2)} = y$

$\dfrac32 \cdot \dfrac{(e^x * x^2)+1}{(e^x * x^2)-1} = y$

11. I see my error now; I though it was just an addition expansion. I get this and now and am happy with the technique. Thanks for your advice.

Regards