Question

If \[a = {\log _2}3\], \[b = {\log _2}5\] and \[c = {\log _7}2\], then \[{\log _{140}}63\] in terms of \[a\], \[b\], \[c\] is
A.\[\dfrac{{2ac + 1}}{{2c + ab + 1}}\]
B.\[\dfrac{{2ac + 1}}{{2a + c + a}}\]
C.\[\dfrac{{2ac + 1}}{{2c + ab + a}}\]
D.None of these

Answer 1

Hint: First, we will start by writing the given logarithm function with same bases using\[{\log _a}b = \dfrac{1}{d}\] for \[{\log _b}a = d\]. Then we will convert the log function to \[{\log _2}\] term. We can do this by using the change-of-base formula, \[{\log _a}x = \dfrac{{{{\log }_b}x}}{{{{\log }_b}a}}\] in the above equation and then use the logarithm property, \[{\log _b}ac = {\log _b}a + {\log _b}c\]and then the power rule of logarithm, \[{\log _b}\left( {{a^c}} \right) = c{\log _b}a\] to simplify the equations to find the required value.

Complete step-by-step answer:
We are given that \[a = {\log _2}3\], \[b = {\log _2}5\] and \[c = {\log _7}2\].
Using the logarithm property, \[{\log _a}b = \dfrac{1}{d}\] for \[{\log _b}a = d\] in \[c = {\log _7}2\], we get
\[ \Rightarrow {\log _2}7 = \dfrac{1}{c}\]
Let us take
\[{\log _{140}}63{\text{ ......eq.(1)}}\]
Rewriting the above equation, we get
\[ \Rightarrow {\log _{{2^2} \times 5 \times 7}}\left( {3 \times 3 \times 7} \right)\]
Let us now start by converting the \[{\log _{{2^2} \times 5 \times 7}}\] to \[{\log _2}\] term. We can do this by using the change-of-base formula, \[{\log _a}x = \dfrac{{{{\log }_b}x}}{{{{\log }_b}a}}\] in the above equation, we get
\[ \Rightarrow \dfrac{{{{\log }_2}\left( {3 \times 3 \times 7} \right)}}{{{{\log }_2}\left( {{2^2} \times 5 \times 7} \right)}}\]
Using the logarithm property, \[{\log _b}ac = {\log _b}a + {\log _b}c\] in the above expression, we get
\[
   \Rightarrow \dfrac{{{{\log }_2}3 + {{\log }_2}3 + {{\log }_2}7}}{{{{\log }_2}4 + {{\log }_2}5 + {{\log }_2}7}} \\
   \Rightarrow \dfrac{{{{\log }_2}3 + {{\log }_2}3 + {{\log }_2}7}}{{{{\log }_2}{2^2} + {{\log }_2}5 + {{\log }_2}7}} \\
 \]
Let us now make use of the power rule of logarithm, \[{\log _b}\left( {{a^c}} \right) = c{\log _b}a\].
So, on applying this rule in the in the above equation, we get
\[ \Rightarrow \dfrac{{{{\log }_2}3 + {{\log }_2}3 + {{\log }_2}7}}{{2{{\log }_2}2 + {{\log }_2}5 + {{\log }_2}7}}\]
Substituting the values \[a = {\log _2}3\], \[b = {\log _2}5\] and \[\dfrac{1}{c} = {\log _2}7\] in the above equation, we get
\[
   \Rightarrow \dfrac{{a + a + \dfrac{1}{c}}}{{2 + b + \dfrac{1}{c}}} \\
   \Rightarrow \dfrac{{2a + \dfrac{1}{c}}}{{2 + b + \dfrac{1}{c}}} \\
   \Rightarrow \dfrac{{\dfrac{{2ac + 1}}{c}}}{{\dfrac{{2c + b + 1}}{c}}} \\
   \Rightarrow \dfrac{{2ac + 1}}{{2c + bc + 1}} \\
 \]
Since none of the options match with the above values, option D is correct.


Note: The power rule can be used for fast exponent calculation using multiplication operation. Students should make use of the appropriate formula of logarithms wherever needed and solve the problem. In mathematics, if the base value in the logarithm function is not written, then the base is \[e\].