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Math Help - simple log question

  1. #1
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    simple log question

    "solve the following equation for x"

    4^(3x-2) = 26^(x+1)

    i use logs and get to :

    (3x-2)log4 = (x+1)log26 ...but not sure where to go from here...simple I imagine.

    thanks
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  2. #2
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    Quote Originally Posted by porge111 View Post
    "solve the following equation for x"

    4^(3x-2) = 26^(x+1)

    i use logs and get to :

    (3x-2)log4 = (x+1)log26 ...but not sure where to go from here...simple I imagine.

    thanks
    "log 4" and "log 26" are just numbers. If it were (3x- 2)A= (x+ 1)B for A and B any numbers, you would multiply to get 3Ax- 2A= Bx+ B. Subtract Bx from both sides and add 2A to both sides to get (3A- B)x= B+ 2A. Finally, (as long as 3A\ne B) divide both sides by 3A- B to get x= \frac{B+ 2A}{3A- B}. Now, replace A by log 4 and replace B by log 26.
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  3. #3
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    4^{3x - 2} = 26^{x + 1}

    \ln{\left(4^{3x - 2}\right)} = \ln{\left(26^{x + 1}\right)}

    (3x - 2)\ln{4} = (x + 1)\ln{(26)}

    (3x - 2)\ln{(2^2)} = (x + 1)\ln{(26)}

    2(3x - 2)\ln{(2)} = (x+1)\ln{(26)}

    \frac{2(3x - 2)}{x + 1} = \frac{\ln{(26)}}{\ln{(2)}}

    \frac{6x - 4}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}

    \frac{6x+6-10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}

    \frac{6x+6}{x+1} - \frac{10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}

    \frac{6(x+1)}{x+1} - \frac{10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}

    6 - \frac{10}{x+1} = \frac{\ln{(26)}}{\ln{(2)}}

    -\frac{10}{x + 1} = \frac{\ln{(26)}}{\ln{(2)}} - 6

    \frac{10}{x+1} = \frac{6\ln{(2)}}{\ln{(2)}} - \frac{\ln{(26)}}{\ln{(2)}}

    \frac{10}{x + 1} = \frac{6\ln{(2)} - \ln{(26)}}{\ln{(2)}}

    \frac{x + 1}{10} = \frac{\ln{(2)}}{6\ln{(2)} - \ln{(26)}}

    x + 1 = \frac{10\ln{(2)}}{6\ln{(2)} - \ln{(26)}}

    x = \frac{10\ln{(2)}}{6\ln{(2)} - \ln{(26)}} - 1

    x = \frac{10\ln{(2)}}{6\ln{(2)} - \ln{(26)}} -  \left(\frac{6\ln{(2)} - \ln{(26)}}{6\ln{(2)} - \ln{(26)}}\right)

    x = \frac{10\ln{(2)} + \ln{(26)} - 6\ln{(2)}}{6\ln{(2)} - \ln{(26)}}

    x = \frac{4\ln{(2)} + \ln{(26)}}{6\ln{(2)} - \ln{(26)}}.
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  4. #4
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    Hello, porge111!

    Solve for x\!:\;\;4^{3x-2} \:=\: 26^{x+1}

    i used logs and get to: . (3x-2)\log4 \:=\:(x+1)\log26

    ... but not sure where to go from here.

    You have: . 3x\!\cdot\!\log4 - 2\!\cdot\!\log 4 \:=\:x\!\cdot\!\log26 + \log26


    . . . . . . . 3x\!\cdot\!\log4 - x\!\cdot\!\log26 \;=\;2\!\cdot\!\log4 + \log26


    Factor: . x\!\cdot\!\bigg(3\log4 - \log26\bigg) \:=\:2\!\cdot\!\log4 + \log26


    Therefore:. . . . . . . . . . . . x \;=\;\dfrac{2\!\cdot\!\log4 + \log26}{3\!\cdot\!\log4 - \log26}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This can be simplified beyond all recognition . . .


    x \;=\;\dfrac{\log(4^2) + \log(26)}{\log(4^3) - \log(26)}

    . . =\;\dfrac{\log(16) + \log(26)}{\log(64) - \log(26)}

    . . =\;\dfrac{\log(16\cdot26)}{\log\left(\frac{64}{26}  \right)}

    . . =\;\dfrac{\log(416)}{\log\left(\frac{32}{13}\right  )}
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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, porge111!


    You have: . 3x\!\cdot\!\log4 - 2\!\cdot\!\log 4 \:=\:x\!\cdot\!\log26 + \log26


    . . . . . . . 3x\!\cdot\!\log4 - x\!\cdot\!\log26 \;=\;2\!\cdot\!\log4 + \log26


    Factor: . x\!\cdot\!\bigg(3\log4 - \log26\bigg) \:=\:2\!\cdot\!\log4 + \log26


    Therefore:. . . . . . . . . . . . x \;=\;\dfrac{2\!\cdot\!\log4 + \log26}{3\!\cdot\!\log4 - \log26}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    This can be simplified beyond all recognition . . .


    x \;=\;\dfrac{\log(4^2) + \log(26)}{\log(4^3) - \log(26)}

    . . =\;\dfrac{\log(16) + \log(26)}{\log(64) - \log(26)}

    . . =\;\dfrac{\log(16\cdot26)}{\log\left(\frac{64}{26}  \right)}

    . . =\;\dfrac{\log(416)}{\log\left(\frac{32}{13}\right  )}
    Or as I put it...

    x = \frac{4\ln{(2)} + \ln{(26)}}{6\ln{(2)}- \ln{(26)}}

    which I prefer since the numbers inside the logarithms are the smallest they can possibly be.
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  6. #6
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    thanks guys!

    I was just being silly as usual it seems.
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