Aptitude - Logarithm - Discussion
Discussion Forum : Logarithm - General Questions (Q.No. 2)
2.
If log 2 = 0.3010 and log 3 = 0.4771, the value of log5 512 is:
Answer: Option
Explanation:
| log5 512 |
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| = 3.876 |
Discussion:
71 comments Page 6 of 8.
Dhivi said:
1 decade ago
Can you say the answer for the (5/512) power 1/9?
Goose said:
1 decade ago
log(75) = log(70+5).
= log(7*10)+(2+3).
= [log(7)+log(10)]*log(2)*log(3).
= [0.8541+1]*0.3010*0.4771.
= 1.8451*0.3010*0.4771.
= 0.1557.
= log(7*10)+(2+3).
= [log(7)+log(10)]*log(2)*log(3).
= [0.8541+1]*0.3010*0.4771.
= 1.8451*0.3010*0.4771.
= 0.1557.
Vipin Mishra said:
1 decade ago
Formula:
log a x = log x/log a.
log 5 512 = log 512/log 5.
log a x = log x/log a.
log 5 512 = log 512/log 5.
Saurabh said:
1 decade ago
Why are they dividing log 512 by log 5?
Raj kumar said:
1 decade ago
@Nirav.
log512/log5 = 9log2/log(2+3) = 9log2/(log2*log3).
= 9*.3010/(.3010*.4771).
= 9/.4771 >18.
log512/log5 = 9log2/log(2+3) = 9log2/(log2*log3).
= 9*.3010/(.3010*.4771).
= 9/.4771 >18.
Nagdish kewat said:
1 decade ago
What is the answer of log3(log2(log3t)) =1 t=?
Nirav said:
1 decade ago
Why are they dividing log 512 by log 5?
Is this a formula?
Is this a formula?
Apoorva said:
1 decade ago
Why is it only 2?
K.rajeshreddy said:
1 decade ago
log5512 = log512/log5 = 9log2/log5.
= 9log2/log3+log2.
= 9*0.3010/0.4771+0.3010.
I think this is this the correct procedure as log 2 and log3 values are given. Why we need to go for log 10/2?
= 9log2/log3+log2.
= 9*0.3010/0.4771+0.3010.
I think this is this the correct procedure as log 2 and log3 values are given. Why we need to go for log 10/2?
Lekan said:
1 decade ago
Please who can solve this log10 2 = 0.3010?
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