Can someone please help me solve this using logarithms?

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

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

Thank you!

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- Sep 23rd 2007, 07:59 AMCoachSolving exponential equations using logarithms
Can someone please help me solve this using logarithms?

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

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

Thank you! - Sep 23rd 2007, 08:08 AMKrizalid
$\displaystyle 5^{4x-1}=7^{x+2}\implies(4x-1)\log5=(x+2)\log7$

Continue. - Sep 23rd 2007, 08:31 AMCoach
I got this far

$\displaystyle x\log_{10}7-4x\log_{10}5=2\log_{10}7-\log_{10}5$ - Sep 23rd 2007, 08:47 AMKrizalid
Carefull with a sign there...

Well, now the next step it is to factorise the LHS by $\displaystyle x$, 'cause that's what we wanna get. - Sep 23rd 2007, 09:16 AMCoach
Thank you, I'll take care of the sign.

yet I don't understand how will factorising help me

$\displaystyle

x({\log_{10}7-3\log_{10}5})=2\log_{10}7-\log_{10}5

$

Do I then just throw everything else on ther side? - Sep 23rd 2007, 09:26 AMKrizalid
Okay, I'm gonna give ya a full solution

$\displaystyle

\begin{aligned}

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

(4x-1)\log5&=(x+2)\log7\\

4x\log5-\log5&=x\log7+2\log7\\

4x\log5-x\log7&=2\log7+\log5\\

\implies x(4\log5-\log7)&=2\log7+\log5\\

\,\therefore\,x&=\frac{2\log7+\log5}{4\log5-\log7}

\end{aligned}

$ - Sep 23rd 2007, 09:58 AMCoach
Thank you!