The value of ${log}_{10} 75 + {log}_{10} 28 - {log}_{10} 21$ is equal to

{log}_{10} 81

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{log}_{10} 20

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $2$
Since the bases of all the terms is same, we can us ${log}_{a} b + {log}_{a} c = {log}_{a} b c$ and ${log}_{a} b - {log}_{a} c = {log}_{a} \frac{b}{c}$
$∴ {log}_{10} 75 + {log}_{10} 28 - {log}_{10} 21$
$= {log}_{10} \frac{75 \times 28}{21} = {log}_{10} 100 = {log}_{10} 10^{2} = 2$

Suggest Corrections

Similar questions

{log}_{10} 5 \cdot {log}_{10} 20 + ({log}_{10} 2)^{2}

{log}_{10} 5. {log}_{10} 20 + ({log}_{10} 2)^{2}

\frac{\cos 3 θ}{2 \cos 2 θ - 1}

\cos θ

\sin θ

\tan θ

\frac{\cos 3 x}{2 \cos 2 x - 1}

For what value of $x$ is $\frac{- 12}{x}$ equal to $3$ ?

Join BYJU'S Learning Program