Rules of Logarithms Combined: Inequality

Is inequality true?

\log_{\frac{1}{4}}9<\frac{\log_57}{\log_5\frac{1}{4}}

Yes, since:

\log_{\frac{1}{4}}9<\log_{\frac{1}{4}}7

$x=\text{?}$

$\ln(x+5)+\ln x≤\ln4+\ln2x$

0 < X \le 3

$x=\text{?}$

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

0 < x\le3.75

$\log_23-\log_2(x+3)\le8$

$x\ge\frac{3}{256}-3$

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

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

\log_{0.25}7+\log_{0.25}\frac{1}{3}<\log_{0.25}x^2

$x=\text{?}$

-\sqrt{\frac{7}{3}} < x < 0,0 < x < \sqrt{\frac{7}{3}}

Find the domain X where the inequality exists

2\log_3x<\log_3(x^2+2x-12)

6 < x

$x=\text{?}$

$\log_{13}(2x^2+3)-\log_{13}2\le\log_{13}7-\log_{13}x^2$

$-\sqrt{2}\le x\le\sqrt{2}$

Given 0<X , find X

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

No solution

Find the domain of X given the following:

\log_{\frac{1}{7}}(x^2+3x)<2\log_{\frac{1}{7}}(3x+1)

No solution

Given X>1 find the domain X where it is satisfied:

$\frac{\log_3(x^2+5x+4)}{\log_3x}<\log_x12$

1 < x < -2.5+\frac{\sqrt{57}}{2}