# Solving exponential equations using logarithms

• Sep 23rd 2007, 07:59 AM
Coach
Solving exponential equations 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 AM
Krizalid
$\displaystyle 5^{4x-1}=7^{x+2}\implies(4x-1)\log5=(x+2)\log7$

Continue.
• Sep 23rd 2007, 08:31 AM
Coach
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 AM
Krizalid
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 AM
Coach
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 AM
Krizalid
Quote:

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

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 AM
Coach
Thank you!