Question
$$2{\log _{10}}^5 + $$ Â $${\log _{10}}8 \,- $$ Â $$\frac{1}{2}{\log _{10}}4$$ Â = ?
Answer: Option A
$$\eqalign{
& 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr
& = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr
& = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr
& = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr
& = {\log _{10}}100 \cr
& = 2 \cr} $$
Was this answer helpful ?
$$\eqalign{
& 2{\log _{10}}5 + {\log _{10}}8 - \frac{1}{2}{\log _{10}}4 \cr
& = {\log _{10}}\left( {{5^2}} \right) + {\log _{10}}8 - {\log _{10}}\left( {{4^{\frac{1}{2}}}} \right) \cr
& = {\log _{10}}25 + {\log _{10}}8 - {\log _{10}}2 \cr
& = {\log _{10}}\left( {\frac{{25 \times 8}}{2}} \right) \cr
& = {\log _{10}}100 \cr
& = 2 \cr} $$
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution