Rules of Logarithms Combined: Using multiple rules

$\log7x+\log(x+1)-\log7=\log2x-\log x$

$?=x$

Defined domain

x>0

x+1>0

x>-1

$\log7x+\log\left(x+1\right)-\log7=\log2x-\log x$

$\log\frac{7x\cdot\left(x+1\right)}{7}=\log\frac{2x}{x}$

We reduce by: $7$ and by $X$

$x\left(x+1\right)=2$

$x^2+x-2=0$

$\left(x+2\right)\left(x-1\right)=0$

$x+2=0$

$x=-2$

Undefined domain x>0

$x-1=0$

$x=1$

Defined domain

$1$

$\log_23x\times\log_58=\log_5a+\log_52a$

Given a>0 , express X by a

$\sqrt[3]{\frac{2a^2}{27}}$

$\log4x+\log2-\log9=\log_24$

?=x

$112.5$

$\log_9e^3\times(\log_224-\log_28)(\ln8+\ln2)$

$6$

$\frac{\log_45+\log_42}{3\log_42}=$

$\log_810$

$\log_64\times\log_9x=(\log_6x^2-\log_6x)(\log_92.5+\log_91.6)$

For all 0 < x

$-3(\frac{\ln4}{\ln5}-\log_57+\frac{1}{\log_65})=$

$3\log_5\frac{7}{24}$

$\frac{\log_76-\log_71.5}{3\log_72}\cdot\frac{1}{\log_{\sqrt{8}}2}=$

$1$

$\frac{1}{\ln4}\cdot\frac{1}{\log_810}=$

$\frac{3}{2}\log e$

Calculate the value of the following expression:

$\ln4\times(\log_7x^7-\log_7x^4-\log_7x^3+\log_2y^4-\log_2y^3-\log_2y)$

$0$

$\frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3=$

$\log_411+\log e^2$

$\frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29=$

$7$

Find X

$\ln8x\times\log_7e^2=2(\log_78+\log_7x^2-\log_7x)$

For all x>0

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

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

Solve for X:

$\ln x+\ln(x+1)-\ln2=3$

$\frac{-1+\sqrt{1+8e^3}}{2}$

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

$-2$

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

$12$

$\log_3x^2\log_527-\log_58=\ln e$

$\pm\sqrt[6]{40}$

$\frac{1}{\log_x3}\times x^2\log_{\frac{1}{x}}27+4x+6=0$

$x=\text{?}$

$\frac{2}{3}+\frac{\sqrt{22}}{3}$

$(2\log_32+\log_3x)\log_23-\log_2x=3x-7$

$x=\text{?}$

$3$