SOLVE FOR X IN TERMS OF LOGA AND LOGB
a^(x-3) = b^(2x)
SO CONFUSED SOMEONE HELP ME!!!
$\displaystyle a^{x-3}=b^{2x}$
Take the log of both sides...
$\displaystyle \log(a^{x-3})=\log(b^{2x})$
The exponents get knocked down outside of the log as coefficients...
$\displaystyle \log(a)(x-3)=\log(b)(2x)$
You can surely solve the rest with a calculator for the logs and algebra for the variable solution!