# Logarithmic Equation

1. Mar 6, 2013

### S.R

1. The problem statement, all variables and given/known data
3^(2x) - 5(3^x) = -6

2. Relevant equations

3. The attempt at a solution
I can't determine the right answer :/? I'm not sure what's incorrect with my current method.

3^(2x) - 5(3^x) = -6

log((3^(2x)) - log((5(3^x)) = -log(6)

xlog(9) - xlog(15)= -log(6)

x((log(9) - log(15)) = -log(6)

x = -log(6)/log(9/15) = 3.507

Any assistance would be appreciated.

2. Mar 6, 2013

### Mentallic

What you essentially did here was:

$$a+b=c$$

$$\log(a+b)=\log(c)$$

$$\log(a)+\log(b)=\log(c)$$

And you can't go from the second to the third line! You can't split up a log that has addition in its argument, and also remember that $$\log(a)+\log(b)=\log(ab)\neq \log(a+b)$$

Start by letting $u=3^x$

EDIT:
Also, $\log(5\cdot 3^x)\neq x\cdot\log(15)$ because to use that rule, it'd have to be $\log((5\cdot 3)^x) = \log(5^x\cdot3^x)$

For $\log(5\cdot 3^x)$ what you'd instead do is convert it into $\log(5)+\log(3^x)$ then apply your rule, $\log(5)+x\cdot\log(3)$

3. Mar 6, 2013

### S.R

Hmm, therefore log(5)+xlog(3) =/= xlog(15)?

Alright, then I can rewrite the equation:

log((3^(2x)) - log((5(3^x)) = -log(6)

xlog(9) - [log(5) + xlog(3)] = -log(6)

x((log(9) - log(3)) = -log(6) + log(5)

x = ((log(5) - log(6)) / log(3)

Correct?

Last edited: Mar 6, 2013
4. Mar 6, 2013

### Dick

No, you can't do any of that for reasons mentallic has already explained. Start from the original equation 3^(2x) - 5(3^x) = -6 and put u=3^x. 3^(2x)=(3^x)^2.

5. Mar 6, 2013

### S.R

Edit: Oh, if I expand, a quadratic equation forms...let me attempt to solve it then substitute u=3^x.

Last edited: Mar 6, 2013
6. Mar 6, 2013

### S.R

Therefore, x=1 and x=log3(2). Thank-you :)

7. Mar 6, 2013

### Dick

Very welcome. You've got it. log3(2) means log to the base 2 of 3, yes? Could also just write it as log(3)/log(2).

Last edited: Mar 6, 2013
8. Mar 6, 2013

### S.R

Yeah :) Additionally, are there any tips you can give for identifying those type of formations/patterns? Or have you simply developed an instinct through experience?

9. Mar 6, 2013

### Dick

Just thinking about it mostly. Having seen a few examples in the past doesn't hurt. So, yeah, let's say experience.

10. Mar 6, 2013

### Mentallic

I'd be lying if I said I could have easily spotted the quadratic equation when first studying logs. When you see a2x in an equation, I always keep in mind that it's equivalent to (ax)2 and thus might involve quadratics.

These are also some of the kinds of problems they use to try and throw you off, because students instinctively think: Well, there are exponentials so I need to use logs, but as you now know, that's not the case.