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Math Help - Multiplication and Subtraction of logs with different bases.

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    Question Multiplication and Subtraction of logs with different bases.

    \text{If }a=\log_{12} 18, b=\log_{24}54\text{, then find the value of }ab+5(a-b)

    I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal.
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    Re: Multiplication and Subtraction of logs with different bases.

    Quote Originally Posted by NameIsHidden View Post
    \text{If }a=\log_{12} 18, b=\log_{24}54\text{, then find the value of }ab+5(a-b)
    I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal.
    Write as $a=\dfrac{\log(2)+2\log(3)}{2\log(2)+\log(3)}$ do the same for $b$.
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    Re: Multiplication and Subtraction of logs with different bases.

    Quote Originally Posted by Plato View Post
    Write as $a=\dfrac{\log(2)+2\log(3)}{2\log(2)+\log(3)}$ do the same for $b$.
    where the "log" on the right can be to any one base. Probably simplest to use base 10.
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    Re: Multiplication and Subtraction of logs with different bases.

    I got $\frac{6(\log 2)(\log 2)+5(\log 2)(\log 3)+6(\log 3)(\log 3)}{6(\log 2)(\log 2)+5(\log 2)(\log 3)+(\log 3)(\log 3)}$
    Is it right saying it $\frac{1}{5(\log 3)(\log 3)}+1$
    Last edited by NameIsHidden; May 30th 2014 at 07:35 AM.
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    Re: Multiplication and Subtraction of logs with different bases.

    Hello, NameIsHidden!

    I'm getting incorrect answers and I don't know why.
    [I know ... I sound like a kid in Algebra I, but I can't help it.]


    \text{If }a\,=\, \log_{12}18,\;b\,=\,\log_{24}54.\;\text{ then find the value of }ab+5(a-b)

    \begin{array}{cccccccc}a \:=\: \log_{12}18 &\Rightarrow& 12^a \:=\: 18 &\Rightarrow& (2^23)^a \:=\: 18 &\Rightarrow& 2^{2a}3^a \:=\: 18 & [1] \\ \\ b \:=\: \log_{24}54 &\Rightarrow&  24^b \:=\: 54 &\Rightarrow&  (2^33)^b \:=\: 54 &\Rightarrow&  2^{3b}3^b \:=\: 54 & [2] \end{array}

    Divide [2] by [1]: . \frac{2^{3b}3^b}{2^{2a}3^a} \:=\:\frac{54}{18} \quad\Rightarrow\quad 2^{3b-2a}3^{b-a} \:=\:3 \:=\:2^03^1

    Equate exponents: . \begin{Bmatrix}3b-2a &=& 0 \\ b-a &=& 1 \end{Bmatrix}

    Solve the system: . \begin{bmatrix}a \:=\:\text{-}3 \\ b \:=\:\text{-}2 \end{bmatrix} \;\; \color{red}{??}
    Thanks from NameIsHidden
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    Re: Multiplication and Subtraction of logs with different bases.

    Is a -3??? No I used a calculator and got 1.163.........
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    Re: Multiplication and Subtraction of logs with different bases.

    Quote Originally Posted by NameIsHidden View Post
    Is a -3??? No I used a calculator and got 1.163.........
    $a=$ This

    And $b=$ this
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    Re: Multiplication and Subtraction of logs with different bases.

    I checked with the solution, the result of the expression is $1$.

    While taking Soroban's explanation, I see that he missed something.

    Actually $a=-3$ or $a=3$ and $b=-2$ or $b=2$.(By substituting two equations in two different ways)

    The final result is either $1$ or $11$.

    Are there any other ways to compute it even accurate?
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