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 1 of 8.
Rohit Saini said:
5 years ago
Logarithm base change rule
The base b logarithm of x is base c logarithm of x divided by the base c logarithm of b.
logb(x) = logc(x) / logc(b)
For example, in order to calculate log2(8) in calculator, we need to change the base to 10:
log2(8) = log10(8) / log10(2)
So by this we get
log5 512 = log 512 / log 5 with base 10
Now since 5 can be written as 10/2
so log 5 = log(10/2)
Now using property log(x / y) = log(x) - log(y)
so, log(10/2) can be written as log(10) - log(2)
So finally we got
log5 512 = log(512) / log(10) - log(2)
since 512 = 2^9
log 512 = log(2^9) = 9 log 2
= 9 log 2 / log 10 - log 2
= (9 x 0.3010) / 1 - 0.3010
= 2.709 / 0.699 = 3.876
Hope it helps !
The base b logarithm of x is base c logarithm of x divided by the base c logarithm of b.
logb(x) = logc(x) / logc(b)
For example, in order to calculate log2(8) in calculator, we need to change the base to 10:
log2(8) = log10(8) / log10(2)
So by this we get
log5 512 = log 512 / log 5 with base 10
Now since 5 can be written as 10/2
so log 5 = log(10/2)
Now using property log(x / y) = log(x) - log(y)
so, log(10/2) can be written as log(10) - log(2)
So finally we got
log5 512 = log(512) / log(10) - log(2)
since 512 = 2^9
log 512 = log(2^9) = 9 log 2
= 9 log 2 / log 10 - log 2
= (9 x 0.3010) / 1 - 0.3010
= 2.709 / 0.699 = 3.876
Hope it helps !
(11)
ArBan said:
9 years ago
How can log 10 be 1? It can only be 1 if the base is 10. If I remember the formula correctly then the base can be anything and is generally a random constant term (a, b etc. ). If we follow the same logic then the answer is most definitely wrong as per my calculations.
CAN ANYONE EXPLAIN THE CORRECT PROCESS?
CAN ANYONE EXPLAIN THE CORRECT PROCESS?
Shalini said:
1 decade ago
@Kasthuri
log 75= log(5*5*3) = log(5*5)+log(3), because log m*n=log m+log n
=2 log5 +log3
=2 log(10/2)+log3
=2(log10-log2)+log3
=2(1-.3010)+..4771
=2*.699+.4771
=1.398+.4771
=1.8751
log 75= log(5*5*3) = log(5*5)+log(3), because log m*n=log m+log n
=2 log5 +log3
=2 log(10/2)+log3
=2(log10-log2)+log3
=2(1-.3010)+..4771
=2*.699+.4771
=1.398+.4771
=1.8751
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?
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.
Sweta said:
7 years ago
@Dinesh Puri.
log 4=log 2^2.
=2log2,
=2log(10/5),
= 2*{log10-log5},
= 2*{1-.690},
= 0.62.
log 4=log 2^2.
=2log2,
=2log(10/5),
= 2*{log10-log5},
= 2*{1-.690},
= 0.62.
Vigneshraj said:
1 decade ago
@jeromy.
Click the impotant formulas link there you can see 7th formula using that formula we divide log512 by log5.
Click the impotant formulas link there you can see 7th formula using that formula we divide log512 by log5.
Girish said:
10 years ago
That is base change rule it seems log (b) to the base a, = log (b) to base c/log (a) to base c i.e log 512/log5.
Xiyaz said:
2 years ago
No @Amrita.
Log 5 can't be written as log 2+log3.
But it can be written as log 5+ log 1.
Hope it helps.
Log 5 can't be written as log 2+log3.
But it can be written as log 5+ log 1.
Hope it helps.
(6)
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.
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