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Thread: Solving Exponential Equations (logs involved)

  1. January 6th 2009, 02:40 PM #1
    mike_302
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    Solving Exponential Equations (logs involved)

    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?
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  2. January 6th 2009, 02:47 PM #2
    chabmgph
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    Quote Originally Posted by mike_302 View Post
    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 assuming this is a system of two equations...

    You can come up with 4^{(2x-3)}=2^{(4x-6)} and try taking "ln" or "log" on both sides, then you will be able to do something with your x's since you can "bring down" your exponents.
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  3. January 6th 2009, 02:50 PM #3
    mike_302
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    No, the two equations are equal... That's as far as I solved. Sorry :P

    And we haven't done ln .

    EDIT: I see what you mean.

    This is what I come up with next.

    3xlog(6)=(2x-3)log(4)

    3xlog(6)=2xlog4-3log4

    log(64)=2xlog(4)-3xlog(6)


    What next :S ?
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  4. January 6th 2009, 02:55 PM #4
    chabmgph
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    Quote Originally Posted by mike_302 View Post
    No, the two equations are equal... That's as far as I solved. Sorry :P

    And we haven't done ln .

    I am not quite sure what you meant by them being equal.

    If say the equation you are trying to solve is 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.

    log(64)=2xlog(4)-3xlog(6)
    well, log(a number) is still a number. so this is similar to solving: 2=2x*5-3x*4<br />

    CAn you take it from here?
    Last edited by chabmgph; January 6th 2009 at 03:05 PM. Reason: OP edit previous post
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  5. January 6th 2009, 03:02 PM #5
    Chop Suey
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    Quote Originally Posted by mike_302 View Post
    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?
    It's really the same idea as the other one you posted. For instance, the first one, if we divide by 4^{2x}, we get

    \frac{6^{3x}}{4^{2x}} = 4^{-3}

    \implies \left(\frac{6^3}{4^2}\right)^x = \frac{1}{64}

    Just try to isolate x, and then solve.
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  6. January 6th 2009, 03:11 PM #6
    Soroban
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    Hello, mike_302!

    Okay, I'll assume you've never seen one of these before.
    I'll give you a walk-through . . .

    Solve for x\!:\;\;6^{3x} \:=\:4^{2x-3}

    Take logs of both sides: . \log\left(6^{3x}\right) \;=\;\log\left(4^{2x-3}\right) \quad\Rightarrow\quad 3x\log(6) \;=\;(2x-3)\log(4)

    . . 3x\log(6) \:=\:2x\log(4) - 3\log(4) \quad\Rightarrow\quad 3x\log(6) - 2x\log(4) \:=\:-3\log(4)

    Factor: . x\bigg[3\log(6) - 2\log(4)\bigg] \:=\:-3\log(4)

    Therefore: . \boxed{x \;=\;\frac{-3\log(4)}{3\log(6) - 2\log(4)}}



    This answer can be simplified beyond all recognition . . .

    \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)}<br />

    . . = \;\frac{\log(64)}{\log(\frac{16}{216})} \;=\;\frac{\log(64)}{\log(\frac{2}{27})} \;=\;\log_{\frac{2}{27}}(64) . . . see what I mean?
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  7. January 6th 2009, 03:11 PM #7
    mike_302
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    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!
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  8. January 6th 2009, 03:15 PM #8
    mike_302
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    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!
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  9. January 7th 2009, 08:14 AM #9
    chabmgph
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    Quote Originally Posted by mike_302 View Post
    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!
    It is common that there are more than one way to approach a math problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in case he/she prefer one than the other, but they are both valid methods.
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