Properties of Logarithms

Solve for $x$: $2+\log x=\log 500$

Prove that $3\log 4-2\log 6+\log {\left(18\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\log \left(96\sqrt{2}\right)$

Solve the equation, ${\log }_3{\left(\frac{x-3}{x-7}\right)}^2+1={\log }_3\left(\frac{x-3}{x-1}\right)$

Prove that $2^{\sqrt{{\log }_{2}3}}=3^{\sqrt{{\log }_{3}2}}$

The value of ${(\log }_{b}a\left)\right({\log }_{c}b)({\log }_{a}c)$ is

Solve: $\log ab-\log \left|b\right|$

The value of $7\mathrm{l}\mathrm{o}\mathrm{g}\frac{16}{15}+5\mathrm{l}\mathrm{o}\mathrm{g}\frac{25}{24}+3\mathrm{l}\mathrm{o}\mathrm{g}\frac{81}{80}$ is equal to

$\frac{{\mathrm{l}\mathrm{o}\mathrm{g}}_{8}17}{{\mathrm{l}\mathrm{o}\mathrm{g}}_{9}23}-\frac{{\mathrm{l}\mathrm{o}\mathrm{g}}_{2\sqrt{2}}17}{{\mathrm{l}\mathrm{o}\mathrm{g}}_{3}23}$, is equal to.

If $a={\log }_{\left(e-1\right)}2+{\log }_2\left(e-1\right)$, then which of the following is correct-

Let $ABC$, be a triangle right-angled at $C$. The value of $\frac{{\mathrm{l}\mathrm{o}\mathrm{g}}_{b+c}a+{\mathrm{l}\mathrm{o}\mathrm{g}}_{c-b}a}{{\mathrm{l}\mathrm{o}\mathrm{g}}_{b+c}a·{\mathrm{l}\mathrm{o}\mathrm{g}}_{c-b}a}$ $\left(b+c\ne 1, c–b\ne 1\right)$, equals (where$a, b, c$, are length of sides opposite to angles $A, B \& C$, respectively).

The value of ${\log }_{3}4.{\log }_{4}5.{\log }_{5}6.{\log }_{6}7.{\log }_{7}8.{\log }_{8}9$ is

The value of  ${\left(\frac{1}{\sqrt{27}}\right)}^{2-\left(\frac{{\log }_{5}16}{{\log }_{5}9}\right)}$  equals to

${49}^{\left(1-{\log }_{7}2\right)}+5^{-{\log }_{5}4}=A$ then the value of $2A$ will be

$4^{{\log }_{2\sqrt{2}}\left(\sqrt{3}-\sqrt{2}\right)+{\log }_{\sqrt{8}}\left(\sqrt{3}+\sqrt{2}\right)}$   is equal to-

If ${\log }_{10}x=y,$ then ${\log }_{1000}{x}^2$ is equal to

The sum of the series ${\log }_{4}2-{\log }_{8}2+{\log }_{16}2\dots ..\infty$, is.

The value of ${\left(0.05\right)}^{{\log }_{\sqrt{20}}\left(0.1+0.01+0.001+\dots \right)}$ is

Solution of equation $4^{\left(\frac{-1}{x}\right)}=\frac{1}{2}\ln \left(\sqrt{e}\right)$ is

Solve: $\log ab-\log \left|b\right|$

$\sum _{n=1}^n\frac{1}{{\log }_{2^{\mathrm{n}}}\left(a\right)}=$