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Math Help - Log equation help

  1. #1
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    Log equation help

    Please can someone take me through the steps in this as I have the x's on both sides

    7^(x-3)=4^(2x)

    The (x-3) and (2x) are exponents

    My math type didnt import the text in

    Thank you
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  2. #2
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    Quote Originally Posted by 200001 View Post
    Please can someone take me through the steps in this as I have the x's on both sides

    7^(x-3)=4^(2x)

    The (x-3) and (2x) are exponents

    My math type didnt import the text in

    Thank you
    7^{x-3} = 4^{2x}

    \log\left(7^{x-3}\right) = \log\left(4^{2x}\right)<br />

    (x-3)\log{7} = 2x\log{4}

    x\log{7} - 3\log{7} = 2x\log{4}

    x\log{7} - 2x\log{4} = 3\log{7}<br />

    x(\log{7} - 2\log{4}) = 3\log{7}

    x = \frac{3\log{7}}{\log{7} - 2\log{4}}
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  3. #3
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    Thank you very much, I see where I went wrong

    What do I do with this one?

    2^x multiplied by 2^(x+1) = 10

    Thanks
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  4. #4
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    Quote Originally Posted by 200001 View Post
    Thank you very much, I see where I went wrong

    What do I do with this one?

    2^x multiplied by 2^(x+1) = 10

    Thanks
    Do you mean 2^x2^{x+1}= 10?
    (The code for that is "2^x2^{x+1}= 10". Notice the "{ }" to group x+ 1 as a single exponent.)

    Use a "law of exponents": 2^x2^{x+1}= 2^{x+ x+ 1}= 2^{2x+1} and then take the logarithm of both sides.
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  5. #5
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    Great, yes I get that

    Now I am on these ones

    9log (base x of 5) = log (base 5 of x)

    not sure how i get through these as Im trying to change the base I believe?
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  6. #6
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    Quote Originally Posted by 200001 View Post
    Great, yes I get that

    Now I am on these ones

    9log (base x of 5) = log (base 5 of x)

    not sure how i get through these as Im trying to change the base I believe?
    in future, start a new problem with a new thread.


    using change of base, note that

    \log_x(5^9) = \frac{\log_5(5^9)}{\log_5{x}} = \frac{9}{\log_5{x}}

    now you have the equation ...

    \frac{9}{\log_5{x}} = \log_5{x}

    9 = (\log_5{x})^2

    \pm 3 = \log_5{x}

    x = 5^3 or x=5^{-3}
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  7. #7
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    Much obliged, thanks
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