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Math Help - Add/Sub Logs of different bases.

  1. #1
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    Add/Sub Logs of different bases.

    Hello,

    Regarding, Solve for x,

    Log(base2)x + Log(base3)x = 1

    I keep getting x = sq root of 6.


    The answer at the back of the book say x = 1.53.

    Is the book wrong? / what am I missing?

    Thank you,
    Gary Og.
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  2. #2
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    Re: Add/Sub Logs of different bases.

    Quote Originally Posted by GaryOg View Post
    Hello,

    Regarding, Solve for x,

    Log(base2)x + Log(base3)x = 1

    I keep getting x = sq root of 6.


    The answer at the back of the book say x = 1.53.

    Is the book wrong? / what am I missing?

    Thank you,
    Gary Og.
    Since I don't know what you did, I can't say where you went wrong.

    I get the exact answer as

    2^{\frac{\log(3)}{\log(6)}} \text{ or }3^{\frac{\log(2)}{\log(6)}}

    I would try by converting everything to a base 10 logarithm

    \frac{\log(x)}{\log(2)}+\frac{\log(x)}{\log(3)}=1 \iff \log x^{\log 3}+x^{\log 2}=\log 2^{\log 3}

    Now try to use the log properties to finish
    Thanks from GaryOg
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  3. #3
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    Re: Add/Sub Logs of different bases.

    Hello, GaryOg!

    \text{Solve for }x\!:\;\;\log_2x + \log_3x \:=\: 1

    I used the Base-Change Formula: . \log_ba \:=\:\frac{\ln a}{\ln b}

    The equation becomes: . \frac{\ln x}{\ln2} + \frac{\ln x}{\ln3} \:=\:1


    Multiply by \ln2\!\cdot\!\ln3\!:

    . . \ln3\!\cdot\!\ln x + \ln2\!\cdot\!\ln x \:=\:\ln2\ln3

    . . . . . . (\ln2 + \ln3)\ln x \:=\:\ln2\!\cdot\!\ln3

    . . . . . . . . . . . . . . \ln x \:=\:\frac{\ln2\!\cdot\!\ln3}{\ln2 + \ln3} \;=\;\frac{\ln2\!\cdot\!\ln3}{\ln6} \;\approx\;0.425

    Therefore: . x \;=\;e^{0.425} \;=\;1.52859... \;\approx\;1.53
    Thanks from GaryOg
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