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:
2.870
2.967
3.876
3.912
Answer: Option
Explanation:
log5 512
= log 512
log 5
= log 29
log (10/2)
= 9 log 2
log 10 - log 2
= (9 x 0.3010)
1 - 0.3010
= 2.709
0.699
= 2709
699
= 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 !
(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?

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

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?

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.

Sweta said:   7 years ago
@Dinesh Puri.

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.

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.
(6)

Raj kumar said:   1 decade ago
@Nirav.

log512/log5 = 9log2/log(2+3) = 9log2/(log2*log3).

= 9*.3010/(.3010*.4771).

= 9/.4771 >18.


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