Exponential Equation

**Slipery** · January 13th 2009, 09:14 AM

Hey guys... I am just finishing up my unit, and I had a question that I am stuck on..

Here it is:

Solve each of the following exponential equations:

b) 4(2^x) + 31 (2^x) - 8 = 0

So, I figure that I can replace (2^x) with just x, and then I would solve it like a complex trinomial.. However, it would then be 4x+31x -8, and I need it to be 4x^2 in order to do that. How am I supposed to solve this?

THanks

**Moo** · January 13th 2009, 09:17 AM

Hey

4X+31X-8=(31+4)X-8=35X-8

Now you can solve the equation 35X-8=0, can't you ? OO
Why are you looking for a trinomial ?

**Slipery** · January 13th 2009, 09:47 AM

Yeah I can,

... Don't know why I am looking for a trinomial.. I guess cause I was solving the previous questions through factoring. If I subsitute x for (2^x), does that mean I need to stick this value back in at some point?

**masters** · January 13th 2009, 09:49 AM

Originally Posted by Slipery

Hey guys... I am just finishing up my unit, and I had a question that I am stuck on..

Here it is:

Solve each of the following exponential equations:

b) 4(2^x) + 31 (2^x) - 8 = 0

So, I figure that I can replace (2^x) with just x, and then I would solve it like a complex trinomial.. However, it would then be 4x+31x -8, and I need it to be 4x^2 in order to do that. How am I supposed to solve this?

THanks

Hello Slipery,

Why even replace $2^x$ ?

Just add the 8 to both sides, factor out the $2^x$ , and then take the log of everything.

$2^x(4+31)=8$

$2^x=\frac{8}{35}$

$\ln 2^x = \ln \frac{8}{35}$

**Slipery** · January 13th 2009, 10:13 AM

Thanks guys!

Masters, I am using your method, and I know how to solve it..

When I am writing it down (just asking cause this is a question that will be marked), I just solve this by changing

to

http://s180.photobucket.com/albums/x...=logquest3.jpg

I am sorry, I do have a rather basic grasp of logs... I am just wondering, how would I show my work to completion using your method of

Because I would just transform this into the diagram I showed above

**masters** · January 13th 2009, 10:22 AM

Originally Posted by Slipery

Thanks guys!

Masters, I am using your method, and I know how to solve it..

When I am writing it down (just asking cause this is a question that will be marked), I just solve this by changing

to

logquest3.jpg - Image - Photobucket - Video and Image Hosting

I am sorry, I do have a rather basic grasp of logs... I am just wondering, how would I show my work to completion using your method of

Because I would just transform this into the diagram I showed above

Hello Slipery,

Review your logarithm rules.

$\ln 2^x = \ln \frac{8}{35}$

$x \ln 2= \ln 8 - \ln 35$

$x=\frac{\ln 8 - \ln 35}{\ln 2}$

$x=$

**Slipery** · January 13th 2009, 10:23 AM

Ahhhh, yes.. I knew that, but I always have a problem remembering when how and which rules to apply =( Thanks alot though

**earboth** · January 13th 2009, 10:32 AM

Originally Posted by Slipery

Hey guys... I am just finishing up my unit, and I had a question that I am stuck on..

Here it is:

Solve each of the following exponential equations:

b) 4(2^x) + 31 (2^x) - 8 = 0

So, I figure that I can replace (2^x) with just x, and then I would solve it like a complex trinomial.. However, it would then be 4x+31x -8, and I need it to be 4x^2 in order to do that. How am I supposed to solve this?

THanks

I don't want to pick at you, but could it be that the original equation reads:

$4 \cdot 2^{2x} + 31\cdot 2^x-8=0$

If and only if: You have a quadratic equation in $2^x$ . Use the substitution $y = 2^x$ . Your equation becomes:

$4y^2+31y-8=0~\implies~ y = \dfrac14~\vee~y=-8$

Now solve

$2^x=\dfrac14 = 2^{-2}~\implies~x = -2$ or

$2^x = -8~\implies~x\notin \mathbb{R}$

**Slipery** · January 13th 2009, 11:21 AM

The original question was correct. You are also correct earboth.. I just didn't know when and when not to use the substitution. The one and only example for the 'substitution' (my book is very limited with examples, and the explaining of the examples), looks the way that your question looked earboth'

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