Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case!
6^(3x)=4^(2x-3)
6^(3x)=2^(4x-6)
We can use the log laws .... And it's likely we do... Anyone give me a hand?
I am not quite sure what you meant by them being equal.
If say the equation you are trying to solve is $\displaystyle 6^{(3x)}=4^{(2x-3)}$, you will need to take log both side. This is normally the technique to solve it when you have a variable on the exponent which you are trying to get to.
EDIT:
I saw you edit your post.
well, log(a number) is still a number. so this is similar to solving: $\displaystyle 2=2x*5-3x*4log(64)=2xlog(4)-3xlog(6)
$
CAn you take it from here?
It's really the same idea as the other one you posted. For instance, the first one, if we divide by $\displaystyle 4^{2x}$, we get
$\displaystyle \frac{6^{3x}}{4^{2x}} = 4^{-3}$
$\displaystyle \implies \left(\frac{6^3}{4^2}\right)^x = \frac{1}{64}$
Just try to isolate x, and then solve.
Hello, mike_302!
Okay, I'll assume you've never seen one of these before.
I'll give you a walk-through . . .
Solve for $\displaystyle x\!:\;\;6^{3x} \:=\:4^{2x-3}$
Take logs of both sides: .$\displaystyle \log\left(6^{3x}\right) \;=\;\log\left(4^{2x-3}\right) \quad\Rightarrow\quad 3x\log(6) \;=\;(2x-3)\log(4) $
. . $\displaystyle 3x\log(6) \:=\:2x\log(4) - 3\log(4) \quad\Rightarrow\quad 3x\log(6) - 2x\log(4) \:=\:-3\log(4) $
Factor: .$\displaystyle x\bigg[3\log(6) - 2\log(4)\bigg] \:=\:-3\log(4) $
Therefore: .$\displaystyle \boxed{x \;=\;\frac{-3\log(4)}{3\log(6) - 2\log(4)}} $
This answer can be simplified beyond all recognition . . .
$\displaystyle \frac{-3\log(4)}{3\log(6) - 2\log(4)} \;=\;\frac{3\log(4)}{2\log(4) - 3\log(6)} \;=\;\frac{\log(4^3)}{\log(4^2) - \log(6^3)} \;=\;\frac{\log(64)}{\log(16)-\log(216)}
$
. . $\displaystyle = \;\frac{\log(64)}{\log(\frac{16}{216})} \;=\;\frac{\log(64)}{\log(\frac{2}{27})} \;=\;\log_{\frac{2}{27}}(64)$ . . . see what I mean?
Okay, just as I was posting, I read Chop Suey's post... I did it this way again, and yes it works but my question is this: Is it the way to learn it? Or is it a quick work around? Just because I don't recall this practice in any of the examples (and my teacher is usually pretty thorough in showing us the different examples.... this looks nothing like any of the ones he gave) .
Thanks for the help!
Lol, I hae to learn to post faster. The posts just roll in!
Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question.
Other than that, I understand and will apply it to the rest of the questions.
Thanks!