Question

If the value of log2 = 0.3010 and log3= 0.4771, then the value of log3.6 is 0.55a2. Where a is the third digit in the expansion. Then the value of a is
[a] 2
[b] 3
[c] 7
[d] 1

Answer 1

Hint: Express 3.6 as a product of powers of 2,3 and 10 only. Use the property log (mn) = log (m) + log (n).
Now use $\log {{a}^{n}}=n\log a$and log 2 = 0.3010, log 3 = 0.4771 and log 10 = 1. Finally, compare the obtained decimal representation with 0.55a2 and find the digit at a place corresponding to “a” in 0.55a2. Hence find the value of a.

Complete step-by-step answer:

We know that \[3.6\text{ }={{2}^{2}}\times {{3}^{2}}\times {{10}^{-1}}\]
Hence,$\log 3.6=\log \left( {{2}^{2}}\times {{3}^{2}}\times {{10}^{-1}} \right)$
Using log (mn) = log (m) +log (n), we get
$\log 3.6=\log \left( {{2}^{2}} \right)+\log \left( {{3}^{2}} \right)+\log \left( {{10}^{-1}} \right)$
Using $\log {{a}^{n}}=n\log a$, we get
$\log 3.6=2\log \left( 2 \right)+2\log \left( 3 \right)-\log \left( 10 \right)$
Using log 2 = 0.3010 , log3 = 0.4771 and log 10 = 1, we get
log(3.6)= 2(0.3010)+2(0.4771)-1
simplifying using BODMAS
=0.6020+0.9552-1
=1.5572-1
=0.5572
Comparing with 0.55a2, we get
a = 7.
Hence a = 7.
Hence option [c] is correct.
Note: The property log mn = log m + log n of logarithms is used to perform large number multiplications.
Suppose , if we want to find the value of mn.
We first find the value of log mn by using log mn = log m + log n
Then we take antilogarithm to get the value of mn.