Change-of-Base Formula for Logarithms: Using variables

$\frac{\log_{4x}9}{\log_{4x}a}=$

$\log_a9$

$\frac{\log_89a}{\log_83a}=$

$\log_{3a}9a$

$\ln4x=$

$\frac{\log_74x}{\log_7e}$

$\frac{2\log_7(x+1)}{\log_7e}=\ln(3x^2+1)$

$x=\text{?}$

$1,0$

$\frac{\log_4(x^2+8x+1)}{\log_48}=2$

$x=\text{?}$

$-4\pm\sqrt{79}$

Find X

$\frac{\log_84x+\log_8(x+2)}{\log_83}=3$

$-1+\frac{\sqrt{31}}{2}$

What is the domain of X so that the following is satisfied:

\frac{\log_{\frac{1}{8}}2x}{\log_{\frac{1}{8}}4}<\log_4(5x-2)

\frac{2}{3} < x

$\frac{\log_47\times\log_{\frac{1}{49}}a}{c\log_4b}=$

$\log_{b^c}\frac{1}{\sqrt{a}}$

$\frac{\log_8x^3}{\log_8x^{1.5}}+\frac{1}{\log_{49}x}\times\log_7x^5=$

$12$

$\log_x16\times\frac{\ln7-\ln x}{\ln4}-\log_x49=$

$-2$

$\frac{\log_x4+\log_x30.25}{x\log_x11}+x=3$

$x=\text{?}$

$2$

$\frac{1}{\log_{2x}6}\times\log_236=\frac{\log_5(x+5)}{\log_52}$

$x=\text{?}$

$1.25$