## Need Some Help With Logarithms: solving solving 2^(2x+1)=3(2^x)-1

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
ZylusC
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### Need Some Help With Logarithms: solving solving 2^(2x+1)=3(2^x)-1

If possible could someone try and talk me through solving 2^(2x+1)=3(2^x)-1

I've tried a number of things to work with it, and ended up finding a block after this first step, which I believed to be:

2(2^(2x))=3(2^x)-1

Any help would be greatly appreciated, as I've lost nearly all of my memory over the summer, after only a week of learning logarithms!
Thank you.

anonmeans
Posts: 84
Joined: Sat Jan 24, 2009 7:18 pm
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### Re: Need Some Help With Logarithms: solving solving 2^(2x+1)=3(2^x)-1

If possible could someone try and talk me through solving 2^(2x+1)=3(2^x)-1

I've tried a number of things to work with it, and ended up finding a block after this first step, which I believed to be:

2(2^(2x))=3(2^x)-1
To get to this point, I think you did this on the left side:

$2^{2x+1}\, =\, \left(2^{2x}\right)\, \left(2^1\right)\, =\, (2^1)\, \left(2^{2x}\right)\, =\, (2)\, \left(2^{2x}\right)\, =\, 2\, \left(2^{2x}\right)$

So now you have this:

$2\, \left(2^{2x}\right)\, =\, 3\, \left(2^{2x}\right)\, -\, 1$

If you use "U" for the bit in the parens, you get:

$2\, U\, =\, 3\, U\, -\, 1$

You should be able to solve this for "U=". (lesson on solving linear eqns) Then back-substitute and solve the exponential equation. (lesson on solving expo. eqns) Please write back if you get stuck.