Rules of Logarithms Combined: Using variables

$x\ln7=$

$\ln7^x$

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

$\log_a9$

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

$\log_{3a}9a$

$(\log_7x)^{-1}=$

$\log_x7$

$n\log_xa=$

$\log_xa^n$

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

$-x^2\log_m3$

$\ln4x=$

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

$\frac{\frac{2x}{\log_89}}{\log_98}=$

$2x$

$\frac{4a^2}{\log_79}\colon\log_97=16$

Calculate a.

$\pm2$

Solve for X:

$\log_3(x+2)\cdot\log_29=4$

$2$

Calculate X:

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

$-4+\sqrt{10}$

$2\log(x+1)=\log(2x^2+8x)$

$x=\text{?}$

$-3+\sqrt{10}$

$\frac{1}{2}\log_3(x^4)=\log_3(3x^2+5x+1)$

$x=\text{?}$

$-\frac{5}{4}\pm\frac{\sqrt{17}}{4}$

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

$x=\text{?}$

$-4\pm\sqrt{79}$

Find X

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

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

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

$x=\text{?}$

$1,0$

$x=\text{?}$

$\log_{\frac{1}{2}}5-\log_{\frac{1}{2}}4\le\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}3$

0 < x\le3.75

$\log_35x\times\log_{\frac{1}{7}}9\ge\log_{\frac{1}{7}}4$

0 < x\le\frac{1}{245}

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

\sqrt{8} < x

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