Find the real values of x such that
3^(2x^2-7x+3)= 4^(x^2-x-6).

Question

Find the real values of x such that 
3^(2x^2-7x+3)= 4^(x^2-x-6).

anonymous · Answer

Tricky . . . going to need a couple steps with logarithms.
Factoring the exponents might help later on, so
$$3^{(2x-1)(x-3)}=4^{(x-3)(x+2)}$$

anonymous · Answer

To make things a little easier to type, I'm going to make the following substitutions:
a=2x-1
b=x-3
c=x+2
so 3^(ab)=4^(bc).
Now take the log, base-3 of both sides.

anonymous · Answer

Can you figure out the next couple steps from there?

anonymous · Answer

how do you take the log of a different base other than 10? because it's been a while since i've worked on logs without a calcultor

anonymous · Answer

Ah, well, unless you want to calculate a decimal approximation, you can leave it in log form.

anonymous · Answer

But in general, let's say you wanted to find the log, base-3 of 4 (which shows up in this problem), you find the log, base-10 of 4 then divide by the log, base-10 of 3. That's the change-of-base formula.

anonymous · Answer

The next step would look like this:
$$\log_3[3^{ab}=4^{bc}] \rightarrow ab=bc \log_3(4).$$

anonymous · Answer

The b's cancel from both sides and you can algebra your way through the rest to solve for x.

anonymous · Answer

I'll chew on  this for a while and get back to you. Logs are not my strong point.

anonymous · Answer

Won't necessarily need this fact for this problem, but remember that logarithms are exponents, so follow all the same rules for exponents.
Good luck!
(The approximate value for x I got was around 4.77)

anonymous · Answer

Wouldn't it be simpler to do log base 10 from both sides to bring the equations down so it will end up looking like:
(ab)(log3) = (bc)(log4) with log base 10?
Or is that an illegal move with logs?