Question

Find the value of n, if $\log 7 - \log 2 + \log 16 - 2\log 3 - \log \dfrac{7}{{45}} = 1 + \log n$.

Answer 1

Hint: In this question we should know the basics to solve this sum so the first term that arises in one mind is what is logarithm. Well in mathematics the logarithm is the inverse function of exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised to produce. By this simple question we can understand the basic build-up of this question as this question revolves around this question. This should make the question easier to solve.

Complete step-by-step answer:

Given,
$
   \Rightarrow \log 7 - \log 2 + \log 16 - 2\log 3 - \log \dfrac{7}{{45}} = 1 + \log n \\
   \Rightarrow \log 7 - \log 2 + \log {2^4} - \log {3^2} - \left( {\log 7 - \log 45} \right) = 1 + \log n \\
 $
$
   \Rightarrow \log 7 - \log 2 + \log {2^4} - \log 9 - \log 7 + \log 45 = 1 + \log n \\
   \Rightarrow \log {2^4} - \log 2 + \log 45 - \log 9 = 1 + \log n \\
 $
$
   \Rightarrow \log \dfrac{{{2^4}}}{2} + \log \dfrac{{45}}{9} = 1 + \log n \\
   \Rightarrow \log \dfrac{{16}}{2} + \log 5 = 1 + \log n \\
   \Rightarrow \log 8 + \log 5 = 1 + \log n \\
   \Rightarrow \log 8 + \log 5 - 1 = \log n \\
   \Rightarrow \log 8 + \log 5 - \log 10 = \log n \\
    \\
 $
$
   \Rightarrow \log 8 + \log \dfrac{5}{{10}} = \log n \\
   \Rightarrow \log 8 + \log \dfrac{1}{2} = \log n \\
   \Rightarrow \log \left( {8 \times \dfrac{1}{2}} \right) = \log n \\
   \Rightarrow \log 4 = \log n \\
  \therefore n = 4 \\
 $

Note: In this question it should be noted that one should know the basics of logarithm. In mathematics, the logarithm is the inverse function of exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. Also note that this solution will be a lot simpler if we use the log values but to make the things simpler I have used the very basic formula to solve the sum for understanding.