Question

# Given that ${{\log }_{2}}3=a,{{\log }_{3}}5=b,{{\log }_{7}}2=c,$ then the logarithm of the number 63 to the base 140 in terms of a, b, c is ?( a )$\dfrac{1+2ac}{2c+abc+1}$ ( b ) $\dfrac{1-2ac}{2c-abc-1}$( c ) $\dfrac{1-2ac}{2c+abc+1}$( d ) $\dfrac{1+2ac}{2c-abc-1}$

Hint: For solving logarithmic related problems, we generally use some of the properties of logarithmic function such as ${{\log }_{N}}M=\dfrac{{{\log }_{a}}M}{{{\log }_{a}}N}$ , ${{\log }_{a}}(M\times N)={{\log }_{a}}M+{{\log }_{a}}N$ , ${{\log }_{a}}M=\dfrac{1}{{{\log }_{M}}a}$, ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$ and so others. Then we will arrange the terms of logarithmic in such a manner so that we can replace the complex log expression to the simplest form using standard results.

In question, we have given that ${{\log }_{2}}3=a$ , ${{\log }_{3}}5=b$ , ${{\log }_{7}}2=c$ , and we have to evaluate the value of ${{\log }_{140}}63$ .
We know that ${{\log }_{N}}M=\dfrac{{{\log }_{a}}M}{{{\log }_{a}}N}$ , ${{\log }_{a}}(M\times N)={{\log }_{a}}M+{{\log }_{a}}N$ and ${{\log }_{a}}M=\dfrac{1}{{{\log }_{M}}a}$ .
Now, we can write ${{\log }_{140}}63$ as $\dfrac{{{\log }_{2}}63}{{{\log }_{2}}140}$ .
Solving further, we get
$\dfrac{{{\log }_{2}}7\times 9}{{{\log }_{2}}5\times 4\times 7}$
Using property, ${{\log }_{a}}(M\times N)={{\log }_{a}}M+{{\log }_{a}}N$ in numerator and denominator both, we get
$\dfrac{{{\log }_{2}}7\times 9}{{{\log }_{2}}5\times 4\times 7}=\dfrac{{{\log }_{2}}7+{{\log }_{2}}9}{{{\log }_{2}}5+{{\log }_{2}}4+{{\log }_{2}}7}$
As, $9={{3}^{2}}$ , we get
$\dfrac{{{\log }_{2}}7+{{\log }_{2}}9}{{{\log }_{2}}5+{{\log }_{2}}4+{{\log }_{2}}7}=\dfrac{{{\log }_{2}}7+{{\log }_{2}}{{3}^{2}}}{{{\log }_{2}}5+{{\log }_{2}}4+{{\log }_{2}}7}$……. ( i )
Now if, ${{\log }_{7}}2=c$ then also $\dfrac{1}{{{\log }_{2}}7}=c$, also ${{\log }_{2}}3=a$ and ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$ then ${{\log }_{2}}{{3}^{2}}=2\cdot {{\log }_{2}}3$,
Putting all these equations in ( i ) , we get
$\dfrac{{{\log }_{2}}7+{{\log }_{2}}{{3}^{2}}}{{{\log }_{2}}5+{{\log }_{2}}4+{{\log }_{2}}7}=\dfrac{\dfrac{1}{c}+2a}{{{\log }_{2}}5+{{\log }_{2}}4+{{\log }_{2}}7}$……. ( ii )
Now, ${{\log }_{2}}5=\dfrac{{{\log }_{3}}5}{{{\log }_{3}}2}$ and further $\dfrac{{{\log }_{3}}5}{{{\log }_{3}}2}={{\log }_{2}}3\cdot {{\log }_{3}}5$ and ${{\log }_{2}}3=a,{{\log }_{3}}5=b,{{\log }_{7}}2=c,$
Putting all these equations in ( ii ) , we get
$\dfrac{\dfrac{1}{c}+2a}{{{\log }_{2}}5+{{\log }_{2}}4+{{\log }_{2}}7}=\dfrac{\dfrac{1}{c}+2a}{a\cdot b+{{\log }_{2}}{{2}^{2}}+\dfrac{1}{c}}$…….. ( iii )
Now, ${{\log }_{2}}{{2}^{2}}=2\cdot {{\log }_{2}}2$ and ${{\log }_{a}}{{a}^{a}}=a$ so ${{\log }_{2}}{{2}^{2}}=2$ .
Putting value of ${{\log }_{2}}{{2}^{2}}$in equation ( iii ) , we get
$\dfrac{\dfrac{1}{c}+2a}{a\cdot b+{{\log }_{2}}{{2}^{2}}+\dfrac{1}{c}}=\dfrac{\dfrac{1}{c}+2a}{a\cdot b+2+\dfrac{1}{c}}$ .
Solving further, we get
$\dfrac{\dfrac{1}{c}+2a}{a\cdot b+2+\dfrac{1}{c}}=\dfrac{\dfrac{1+2ac}{c}}{\dfrac{abc+2c+1}{c}}$ .
On simplifying, we get
$\dfrac{\dfrac{1+2ac}{c}}{\dfrac{abc+2c+1}{c}}=\dfrac{1+2ac}{1+abc+2c}$ .

So, the correct answer is “Option A”.

Note: Firstly, we must know all the properties of logarithmic functions such as ${{\log }_{N}}M=\dfrac{{{\log }_{a}}M}{{{\log }_{a}}N}$ , ${{\log }_{a}}(M\times N)={{\log }_{a}}M+{{\log }_{a}}N$ , ${{\log }_{a}}M=\dfrac{1}{{{\log }_{M}}a}$, ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$. One of the most basic tricks to solve logarithmic functions is to re – arrange the order of base value and variable value such that you left with some expression to which you can replace with some standard results.