# Solving exponential equations using logarithms

• Sep 23rd 2007, 08:59 AM
Coach
Solving exponential equations using logarithms

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

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

Thank you!
• Sep 23rd 2007, 09:08 AM
Krizalid
$5^{4x-1}=7^{x+2}\implies(4x-1)\log5=(x+2)\log7$

Continue.
• Sep 23rd 2007, 09:31 AM
Coach
I got this far

$x\log_{10}7-4x\log_{10}5=2\log_{10}7-\log_{10}5$
• Sep 23rd 2007, 09:47 AM
Krizalid
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.
• Sep 23rd 2007, 10:16 AM
Coach
Thank you, I'll take care of the sign.

yet I don't understand how will factorising help me

$
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, 10:26 AM
Krizalid
Quote:

Originally Posted by Coach
$5^{4x-1}=7^{x+2}$

Okay, I'm gonna give ya a full solution


\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, 10:58 AM
Coach
Thank you!