The value of $\log _{2} 16$ is

$2 \log _{10} 5 + \log _{10} 8 - \frac{1}{2} \log _{10} 4 = ?$

If $\log_{10} 2 = 0.3010$ and $\log_{10} 3 = 0.4771$,then the number of zeros between the decimal point and the first significant figure in $(0.0432)^{10}$ is:

If $\log _{10} 5 + \log _{10}(5x + 1) = \log _{10}(x + 5) + 1$,then $x$ is equal to

For $x>0,$ if $y=\frac{10^{\log _{10} x}}{x^{2}}$ and $x=y^{a},$ then $a=$

If $\log _{10} 2986 = 3.4751$,then $\log _{10} 0.02986 =$