Power Property of Logorithms: Applying the formula

$2\log_38=$

$\log_364$

$3\log_76=$

$\log_7216$

$\log_68=$

$3\log_62$

$x\ln7=$

$\ln7^x$

$n\log_xa=$

$\log_xa^n$

$x\log_m\frac{1}{3^x}=$

$-x^2\log_m3$

Calculate X:

$2\log(x+4)=1$

$-4+\sqrt{10}$

\log_{\frac{1}{3}}e^2\ln x<3\log_{\frac{1}{3}}2

\sqrt{8} < x

Given 0<X , find X

$\log_4x\times\log_564\ge\log_5(x^3+x^2+x+1)$

No solution