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December 20, 2014

December 20, 2014

Posted by **Ang** on Friday, October 26, 2007 at 10:59pm.

4^(x + 4) = 5^(2x + 5)-This question I dont understand.

- exponential equation-Algebra -
**Corin**, Friday, October 26, 2007 at 11:37pmAlright. I think I might have gotten the answer. Write back and tell me if you think it's right. If you think I messed up, I'll keep trying.

First use ln to rearrange the original:

4^(x+4)=5^(2x+5)

ln4^(x+4)=ln5^(2x+5)

Put the exponents in front because of the properties of logs.

(x+4)(ln4)=(2x+5)(ln5)

Rearrange to get all the x's on one side.

(x+4)/(2x+5)=(ln5)/(ln4)

I'm sure you got to this point. But after here, it starts to get a little tricky.

Technically, (ln5)/(ln4) is equal to [(ln5)/(ln4)]/1 right?

Let's set a variable......how about "u" equal to the numerator of this fraction.

So... u=[(ln5)/(ln4)]

Let's plug that back in to what we had from before:

(x+4)/(2x+5)=u/1

Use cross multlipication to get this:

(x+4)=(2x+5)(u)

Distribute the u

x+4=2xu+5u

Get all the parts with x's on one side:

x-2xu=5u-4

Pull out an x:

x(1-2u)=5u-4

Divide the (5u-4) by (1-2u) to get x by itself:

x=(5u-4)/(1-2u)

Plug the u value back in and simplify

x=(5((ln5)/(ln4))-4) / (1-2((ln5)/(ln4))

I think that's as far as you can go.

Good luck! :-)

- exponential equation-Algebra -
**Ang**, Friday, October 26, 2007 at 11:52pmthe closest answers that I have that are given in the multiple choice is {5 ln 5 - 4 ln 4/ln 4-2 ln 5}

- exponential equation-Algebra -
**drwls**, Saturday, October 27, 2007 at 2:14amTake the natural logs (ln) of both sides.

(x+4) ln 4 = (2x+5) ln 5

x ln 4 + 4 ln 4 = 2x ln 5 + 5 ln 5

x (2 ln 5 - ln 4) = 4 ln 4 - 5 ln 5

x = (4ln4 - 5ln5)/(2ln5 - ln4)

This becomes the same as your answer if you multiply both numerator and denominiator by -1

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