Evaluate:
${log}_{10} 1000 + {log}_{9} 81$

Open in App

Solution

${log}_{10} 1000 + {log}_{9} 81$

$= {log}_{10} 10^{3} + {log}_{9} 9^{2}$

$= 3 {log}_{10} 10 + 2 {log}_{9} 9$

$= 3 + 2 = 5$

Suggest Corrections

Similar questions

{log}_{10} 1000 + {log}_{9} 729

{log}_{9} 729 + {log}_{9} 81 + {log}_{9} 9

{log}_{9} 9 + {log}_{9} 81 + {log}_{9} 729

{log}_{10} 1000

Evaluate : $\int \frac{x^{3}}{(x - 1) (x^{2} + 1)} d x .$

Evaluate $\int \frac{s i n x - x c o s x}{x (x + s i n x)}$ dx.

Join BYJU'S Learning Program