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Math Help - Please help from this simple exponential equation problem!

  1. #1
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    Unhappy Please help from this simple exponential equation problem!

    I'm basically done all the questions that I needed to do, yet this one I can't seem to find how to answer for some reason.

    3(2)^x = 4^(x + 1)

    To find x, I try to put it to logarithmic function,

    3*(x)(Log 2) = (x + 1)(Log 4)

    My question is, is that where the 3 goes? It multiplies (x)(Log 2)? Doing it this way I end up with,

    0.9x - 0.6x = 0.6
    x = 2

    The correct answer however is -0.42

    I really need help what to do with that 3, as I've read the book plenty now and I still don't know how this works! Please help!

    There's also another question similar to this, and of course I can't answer it either.

    5(4)^2x = 4(2)^6x

    Any help is greatly appreciated!
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  2. #2
    Senior Member eumyang's Avatar
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    Remember that 4 = 2^2, so you could rewrite like this:
    \begin{aligned}<br />
3 \cdot 2^x &= 4^{x + 1} \\<br />
3 \cdot 2^x &= (2^2)^{x + 1} \\<br />
3 \cdot 2^x &= 2^{2x + 2} \\<br />
\end{aligned}
    Can you take it from here?

    Regarding the second,
    \begin{aligned}<br />
5 \cdot 4^{2x} &= 4 \cdot 2^{6x} \\<br />
5 \cdot (2^2)^{2x} &= 2^2 \cdot 2^{6x} \\<br />
5 \cdot 2^{4x} &= 2^{6x + 2} \\<br />
 \end{aligned}
    Can you take it from here?
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  3. #3
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    Quote Originally Posted by eumyang View Post
    Remember that 4 = 2^2, so you could rewrite like this:
    \begin{aligned}<br />
3 \cdot 2^x &= 4^{x + 1} \\<br />
3 \cdot 2^x &= (2^2)^{x + 1} \\<br />
3 \cdot 2^x &= 2^{2x + 2} \\<br />
\end{aligned}
    Can you take it from here?

    Regarding the second,
    \begin{aligned}<br />
5 \cdot 4^{2x} &= 4 \cdot 2^{6x} \\<br />
5 \cdot (2^2)^{2x} &= 2^2 \cdot 2^{6x} \\<br />
5 \cdot 2^{4x} &= 2^{6x + 2} \\<br />
 \end{aligned}
    Can you take it from here?
    Hi, thanks for the reply! However I'm still not sure how to solve for x. I'm suppose to use Logarithmic function, I just don't know what to do with "3" when I use Logarithmic function on 3(2)^x = 4^(x + 1)
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  4. #4
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    3 \cdot 2^x &= 2^{2x + 2}

    3  &= \frac{2^{2x + 2}}{2^x}

    3  &= 2^{2x + 2-x}

    3  &= 2^{x + 2}

    \log_23  &= {x + 2}

    Now can you finish?
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  5. #5
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    Quote Originally Posted by pickslides View Post
    3 \cdot 2^x &= 2^{2x + 2}

    3  &= \frac{2^{2x + 2}}{2^x}

    3  &= 2^{2x + 2-x}

    3  &= 2^{x + 2}

    \log_23  &= {x + 2}

    Now can you finish?
    Yes this helped! Well I had to still figure out how it works, but looking at this helped me still.

    3*(4)^x also means Log 3 + (x) Log 4. Thanks very much for the replies, very much appreciated =)
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  6. #6
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    Quote Originally Posted by garougt View Post
    Well I had to still figure out how it works,
    You need to remember that a^b = c\implies b = \log_ac
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