Solve Logarithmic Equation: 3^(2x) - 5(3^x) = -6

[S.R]

Homework Statement

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

Homework Equations

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.

 

[Mentallic]

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

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

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, $\log(5)+x\cdot\log(3)$

 

[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?

 

[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.

 

[S.R]

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

 

[S.R]

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

 

[Dick]

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

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).

 

[S.R]

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).

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?

 

[Dick]

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?

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

 

[Mentallic]

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?

I'd be lying if I said I could have easily spotted the quadratic equation when first studying logs. When you see a^2x in an equation, I always keep in mind that it's equivalent to (a^x)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.