Question

Given that ${{\log }_{5}}a\cdot {{\log }_{a}}x=2,$ then the value of x is equals to ?
( a ) 125
( b ) ${{a}^{2}}$
( c ) 25
( d ) none of these

Answer 1

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 logarithmic expression to the simplest form using standard results .

Complete step-by-step solution:
In question, we have given that ${{\log }_{5}}a\cdot {{\log }_{a}}x=2,$ and we have to evaluate the value of x .
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 ${{\log }_{5}}a\cdot {{\log }_{a}}x=2,$ as $\dfrac{{{\log }_{2}}a}{{{\log }_{2}}5}\cdot \dfrac{{{\log }_{2}}x}{{{\log }_{2}}a}=2$.
Solving further, we get
$\dfrac{{{\log }_{2}}x}{{{\log }_{2}}5}=2$
Now we know that ${{\log }_{a}}{{a}^{a}}=a$, so we can write 2 as ${{\log }_{2}}{{2}^{2}}=2$ .
$\dfrac{{{\log }_{2}}x}{{{\log }_{2}}5}={{\log }_{2}}{{2}^{2}}$
Taking ${{\log }_{2}}5$from denominator of fraction on left hand side to numerator on right side, we get
${{\log }_{2}}x={{\log }_{2}}{{2}^{2}}\cdot {{\log }_{2}}5$
On solving, we get,
${{\log }_{2}}x=2\cdot {{\log }_{2}}5$
Using property of logarithmic stated as, ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$ .
So, we can write $2\cdot {{\log }_{2}}5$ as ${{\log }_{2}}{{5}^{2}}$
${{\log }_{2}}x={{\log }_{2}}{{5}^{2}}$
Now, on simplifying we get,
${{\log }_{2}}x={{\log }_{2}}25$
Whenever we have an expression as ${{\log }_{c}}a={{\log }_{c}}b$, then this states that a = b, which means the logarithmic of two numbers to the same base leads same value if and only if both numbers are same.
Hence, ${{\log }_{2}}x={{\log }_{2}}25$
$x=25$

Hence, the option ( c ) is correct.

Note: Firstly, we must know all the properties of logarithmic functions such as ${{\log }_{N}}M=\dfrac{{{\log }_{a}}M}{{{\log }_{a}}N}$ , ${{\log }_{a}}M\times N={{\log }_{a}}M+{{\log }_{a}}N$ , ${{\log }_{a}}M=\dfrac{1}{{{\log }_{M}}a}$, ${{\log }_{a}}{{M}^{r}}=r.{{\log }_{a}}M$, ${{\log }_{a}}{{a}^{a}}=a$. The logarithmic of two numbers to the same base leads same value if and only if both numbers are same .One of the most basic trick 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.