## 3^x = 12x-9

1. The problem statement, all variables and given/known data

3^x=12x-9

2. Relevant equations

3. The attempt at a solution

I really have no clue how to solve this one algebraically.
I graphed the two functions on a calculator and found the points of intersection
the answers are 3 and 1

Can someone show me how to solve this problem algebraically, step by step?

I believe you use logs?

thanks
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Homework Help
 Quote by physicsdreams Can someone show me how to solve this problem algebraically, step by step?
No, we're not allowed to do that. That's against forum rules.

I'm not sure it can be solved algebraically. The solution provided by Wolfram used the Lambert W-function, but I have to confess that I don't remember what that is.
 $$3^x=3(4x-3)$$ When does the left side equal the right side?

Recognitions:
Homework Help

## 3^x = 12x-9

 Quote by eumyang No, we're not allowed to do that. That's against forum rules. I'm not sure it can be solved algebraically. The solution provided by Wolfram used the Lambert W-function, but I have to confess that I don't remember what that is.
The function f(x)=xe^x is a one-to-one function if you restrict the domain to x in [-1,infinity). So it has an inverse. The inverse is the Lambert W-function. It's just a name for a function you can't express algebraically.
 thanks guys for the help, sorry about asking for the step by step, I'm new here. as for the Lambert w-function, is it something one learns in Calculus or something? I'm only in precalculus.

Recognitions:
Homework Help
 Quote by physicsdreams thanks guys for the help, sorry about asking for the step by step, I'm new here. as for the Lambert w-function, is it something one learns in Calculus or something? I'm only in precalculus.
I think I learned about it in a course called Mathematical Physics which is postcalculus, where you study all sorts of special functions associated with differential equations. I don't think are supposed to solve the equation like that. Do just what you did. Sketch the graph and guess the roots.
 Blog Entries: 1 You can solve this algebraically.
 Blog Entries: 1 $$3^x = 3(4x-3)$$ $$3^{x-1} = 4x-3$$ $$3^{x-1} + 3 = 4x$$ $$3^{x-2} + 1 = \frac{4x}{3}$$ $$\frac{3^{x-2}}{x} + \frac{1}{x} = \frac{4}{3}$$ See if you can go from there.

Mentor
 Quote by mjordan2nd $$3^x = 3(4x-3)$$ $$3^{x-1} = 4x-3$$ $$3^{x-1} + 3 = 4x$$ $$3^{x-2} + 1 = \frac{4x}{3}$$ $$\frac{3^{x-2}}{x} + \frac{1}{x} = \frac{4}{3}$$ See if you can go from there.
And this is helpful...how?

 Quote by Mark44 And this is helpful...how?
I am assuming that it is, in fact, impossible to solve algebraically?
I really don't see how moving everything around gets me closer to an answer.

Thanks for trying though, mjordan2nd, unless there really is a way to solve it by taking your route. Please feel free to give me a few more hints as to how this helps.

(I'm still learning here!)

Thanks

Blog Entries: 1
 Quote by Mark44 And this is helpful...how?
Ahh, I suppose you're right. My thinking was you could simply equate the top with the top and the bottom with the bottom. In this case equating the bottom with the bottom gives x = 3, which happens to work in this case, but doesn't have to. My fault.

 Tags algebra, logs