Question

If a = log 15 and b = log 50, then ${\log _9}40$ is equal to:
A) $\dfrac{{5 - 2b}}{{2(a - b + 1)}}$
B) $\dfrac{{b - 1}}{{2(a - b + 1)}}$
C) $\dfrac{{b - 1}}{{2(a - b + 1)}}$
D) $\dfrac{{5 - 2b}}{{(a - b + 1)}}$

Answer 1

In order to solve the question we take the hit and trial method by using the given data $a = \log 15$ and$b = \log 50$on each option. We will consider $\log 10 = 1$ whenever we need it.

Complete step by step answer:
Given:-
$a = \log 15$and $b = \log 50$

Calculating the value for the term:
$(a - b + 1)$…………. (i)
Substituting the given values of a and b in eqn (i), we get.
$a - b + 1 = \log 15 - \log 50 + \log 10$
$\Rightarrow a - b + 1 = \log \dfrac{{15 \times 10}}{{50}}$
$\Rightarrow a - b + 1 = \log 3$……………..(ii)
Multiplying eqn (ii) by
$\Rightarrow 2(a - b + 1) = 2\log 3$
Using: $\log {m^n} = n\log m$
$\Rightarrow 2(a - b + 1) = \log 9$
Therefore, $2(a - b + 1) = \log 9.$…………… (ii)
Now, calculating the value for term:
$(5 - 2b)$.......................... (iii)
Substituting the given values of a and b in eqn (iii), we get:
$5 - 2b = 5\log 10 - 2 \times \log 50$…………………… (iv)
$= \log \dfrac{{{{10}^5}}}{{{{50}^2}}}$
$= \log \dfrac{{100 \times 100 \times 10}}{{50 \times 50}}$
$= \log 40.$
Therefore we finally have
$5 - 2b = \log 40.$
Dividing eqn (iv) from eqn (ii) we get:
\Rightarrow $$$$\dfrac{{5 - 2b}}{{2(a - b + 1)}} = \dfrac{{\log 40}}{{\log 9}}

Hence, option (A) is the correct answer.

Note:
Sometimes questions can be solved by elimination process right in this question, we should be well known about the concept of the logarithm. In this question, some formulas have been used to solve this question.

$$\log \left( {m \times n} \right) = \log m + \log n \\\ \log {m^n} = n\log m \\\$$