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Math Help - Solving exponential equations using logarithms

  1. #1
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    Solving exponential equations using logarithms

    Can someone please help me solve this using logarithms?

    5^{4x-1}=7^{x+2}


    I made them into logarithms with the base of ten, but after that I got stuck.









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  2. #2
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    5^{4x-1}=7^{x+2}\implies(4x-1)\log5=(x+2)\log7

    Continue.
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  3. #3
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    I got this far

    x\log_{10}7-4x\log_{10}5=2\log_{10}7-\log_{10}5
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  4. #4
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    Carefull with a sign there...

    Well, now the next step it is to factorise the LHS by x, 'cause that's what we wanna get.
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  5. #5
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    Thank you, I'll take care of the sign.

    yet I don't understand how will factorising help me

     <br />
x({\log_{10}7-3\log_{10}5})=2\log_{10}7-\log_{10}5<br />










    Do I then just throw everything else on ther side?
    Last edited by Coach; September 23rd 2007 at 09:39 AM.
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  6. #6
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    Quote Originally Posted by Coach View Post
    5^{4x-1}=7^{x+2}
    Okay, I'm gonna give ya a full solution

     <br />
\begin{aligned}<br />
5^{4x-1}&=7^{x+2}\\<br />
(4x-1)\log5&=(x+2)\log7\\<br />
4x\log5-\log5&=x\log7+2\log7\\<br />
4x\log5-x\log7&=2\log7+\log5\\<br />
\implies x(4\log5-\log7)&=2\log7+\log5\\<br />
\,\therefore\,x&=\frac{2\log7+\log5}{4\log5-\log7}<br />
\end{aligned}<br />
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  7. #7
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    Thank you!
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