The value of ${log}_{10} (3 / 2) + {log}_{10} (4 / 3) + . . . .$ up to 99 terns

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

2.5

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

None of the above

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

Open in App

Solution

The correct option is C None of the above
Using
$log a$ + $log b$ = $log a b$
= ${log}_{10} (\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times . . . . . . . \times \frac{101}{100})$
= ${log}_{10} \frac{101}{2}$

Suggest Corrections

Similar questions

{log}_{10} (\frac{3}{2}) + {log}_{10} (\frac{4}{3}) + {log}_{10} (\frac{5}{4}) + . . . .

8

{log}_{10} 4 = 0.6020

{log}_{10} 8

{cos}^{- 1} \frac{x - 3}{2} - {log}_{10} (4 - x)

\leq

[\frac{2}{3}] + [(\frac{2}{3}) + (\frac{1}{99})] + [(\frac{2}{3}) + (\frac{2}{99})] + \dots \dots + [(\frac{2}{3}) + \frac{98}{99}]

{log}_{10} 4 = 0.6021

{log}_{10} 5 = 0.6990

{log}_{10} 1600

Join BYJU'S Learning Program