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!
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?
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}
$