Question

The value of \({\log _{\dfrac{1}{{20}}}}40\) is
A. Greater than zero
B. Smaller than zero
C. Greater than zero and smaller than one
D. None of these

Answer 1

Hint: To solve the question given above, we will find out what is the definition of logarithm function. Then, we will look at the general form of logarithm function i.e. \({\log _a}b\).Then, we will use the identity: \({\log _{{a^2}}}b = \dfrac{1}{2}{\log _a}b\) to convert the given logarithm function into other terms. Then, we will determine in which range, the value of the given logarithm function.

Complete step-by-step answer:
Before we solve the question given above, we will first see what a logarithm function is. A logarithm is the power in which a number must be raised in order to get some other number. The logarithm is the inverse function to exponentiation. The general form of logarithm is \({\log _a}b\) .Now, we are given a logarithm. We will assume that its value is x. Thus,
\(x = {\log _{\dfrac{1}{{20}}}}40{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (i)}}\)
Now, we will use an exponential property here, which says that \({a^{ - 1}} = \dfrac{1}{a}\) .Thus, we get:
\(x = {\log _{{{20}^{ - 1}}}}40\)
Now, we will use a logarithm identity in the above question:
\({\log _{{a^2}}}b = \dfrac{1}{2}{\log _a}b\)
Thus, we get:
\(\begin{array}{l}x = - {\log _{20}}40\\x = - {\log _{20}}\left( {20 \times 2} \right)\end{array}\)
Now, we will use \(\log \left( {a \times b} \right) = \log a + \log b\) in above equation. Thus, we will get:
\(\begin{array}{l}x = - \left[ {{{\log }_{20}}20 + {{\log }_{20}}2} \right]\\x = - \left[ {1 + lo{g_{20}}2} \right]\end{array}\)
Now, we will use \({\log _a}b = \dfrac{1}{{{{\log }_b}a}}\) in above question:
\[\begin{array}{l}x = - \left[ {1 + \dfrac{1}{{lo{g_2}20}}} \right]\\ \Rightarrow x = - \left[ {1 + \dfrac{1}{{lo{g_2}\left( {4 \times 5} \right)}}} \right]\\ \Rightarrow x = - \left[ {1 + \dfrac{1}{{lo{g_2}4 + lo{g_2}5}}} \right]\\ \Rightarrow x = - \left[ {1 + \dfrac{1}{{2 + lo{g_2}5}}} \right]{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (ii)}}\end{array}\]
Now, we consider the value of \({\log _2}5\) as y. Thus,
\(y = {\log _2}5{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (iii)}}\)
We use the identity: \({\log _a}b = \dfrac{{{{\log }_{10}}b}}{{{{\log }_{10}}a}}\). Thus, we get:
\(\begin{array}{l} \Rightarrow y = \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}2}}\\ \Rightarrow y = \dfrac{{{{\log }_{10}}\left( {\dfrac{{10}}{2}} \right)}}{{{{\log }_{10}}2}}\end{array}\)
Now, we will use the identity: \(\log \left( {\dfrac{a}{b}} \right) = \log a - \log b\)
\(\begin{array}{l} \Rightarrow y = \dfrac{{{{\log }_{10}}10 - {{\log }_{10}}2}}{{{{\log }_{10}}2}}\\ \Rightarrow y = \dfrac{{1 - {{\log }_{10}}2}}{{{{\log }_{10}}2}}\end{array}\)
The value of \({\log _{10}}2\) is 0.3010. Thus, we have:
\(\begin{array}{l} \Rightarrow y = \dfrac{{1 - 0.3010}}{{0.3010}}\\ \Rightarrow y = \dfrac{{0.6990}}{{0.3010}}\\ \Rightarrow y = 2.322{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (iv)}}\end{array}\)
From (iii) and (iv), we have:
\({\log _2}5 = 2.322{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (v)}}\)
From (ii) and (v), we have the following equation:
\(\begin{array}{l} \Rightarrow x = - \left[ {1 + \dfrac{1}{{2 + 2.322}}} \right]\\ \Rightarrow x = - \left[ {1 + \dfrac{1}{{4.322}}} \right]\\ \Rightarrow x = - \left[ {1 + 0.231} \right]\\ \Rightarrow x = - \left[ {1.231} \right]\\ \Rightarrow x = - 1.231{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (vi)}}\end{array}\)
From (i) and (vi), we can say that:
\(\begin{array}{l}{\log _{\dfrac{1}{{20}}}}40 = - 1.231\\ \Rightarrow {\log _{\dfrac{1}{{20}}}}40 < 0\end{array}\)
Hence, option B is correct.

Note: The alternate method to do this question is shown below.
Let the logarithm be \({\log _a}b\). If a>1 and b>1 then the value of logarithm will always be positive, if a>1 and b<1 then the value of logarithm will always be negative. If a<1 and b<1 then the value of logarithm is positive and if a<1 and b>1 then logarithm gives negative value. In our case a<1 and b>1. So the value of the logarithm will be less than 0.