If log72 = m, then log4928 is equal to ? 1/(1+2m) (1+2m)/2 2m/(2m+1) (2m+1)/2m l

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Question
If log72 = m, then log4928 is equal to ?
Options
A. 1/(1+2m)
B. (1+2m)/2
C. 2m/(2m+1)
D. (2m+1)/2m

Correct Answer

(1+2m)/2

Explanation

$\log_{49} (28) = \frac{1}{2 \log_{7} (7 � 4)}$

= $\frac{1}{2} + \frac{1}{2} (2 \log_{7} (2))$
= $\frac{1}{2} + \log_{7} (2)$
= $\frac{1}{2} + m$
= $\frac{1 + 2 m}{2}$ .

Logarithm problems

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3. If log 27 = 1.431, then the value of log 9 is
Options
A. 0.754
B. 0.854
C. 0.954
D. 0.654

4. If log 2 = 0.30103, Find the number of digits in 256 is
Options
A. 17
B. 19
C. 23
D. 25

5. If log 2 = 0.3010 and log 3 = 0.4771, the values of log5 512 is
Options
A. 2.875
B. 3.875
C. 4.875
D. 5.875

6. Find the logarithm of 144 to the base 23 :
Options
A. 2
B. 4
C. 8
D. None of these

8. If log 64 = 1.8061, then the value of log 16 will be (approx)?
Options
A. 1.9048
B. 1.2040
C. 0.9840
D. 1.4521

9. A fast moving superfast express crosses another pasenger train in 20 seconds. The speed of faster train is 72 km/hr and speeds of slower train is 27 km/h. Also the length of faster ntrain is 100m, then find the length of the slower train if they are moving in the same direction.
Options
A. 100 m
B. 125 m
C. 150 m
D. 175 m

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