Question

Find the value of ${{a}^{\dfrac{{{\log }_{b}}{{\log }_{b}}N}{{{\log }_{b}}a}}}$ is; \[\]
A.${{\log }_{b}}N$ \[\]
B. $-{{\log }_{b}}N$ \[\]
C. ${{\log }_{N}}b$ \[\]
D. ${{\log }_{N}}b$ \[\]

Answer 1

Hint: We use the base change formula ${{\log }_{b}}x=\dfrac{{{\log }_{d}}x}{{{\log }_{d}}b}$ to change the given base of the in the numerator and denominator of the logarithmic fraction present in the logarithmic fraction\[\dfrac{{{\log }_{b}}{{\log }_{b}}N}{{{\log }_{b}}a}\]. We simplify and then use the formula ${{b}^{{{\log }_{b}}x}}=x$ to get the answer. \[\]

Complete step by step answer:
We know that the logarithm is the inverse operation to 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$x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
So we can write ${{b}^{y}}=x$ as
\[\begin{align}
  & {{b}^{y}}=x \\
 & \Rightarrow {{b}^{{{\log }_{b}}x}}=x....\left( 1 \right) \\
\end{align}\]
We change the base $b$ of the logarithm to base $d$ using the following formula
\[{{\log }_{b}}x=\dfrac{{{\log }_{d}}x}{{{\log }_{d}}b}......\left( 2 \right)\]
We are asked to find the value of ${{a}^{\dfrac{{{\log }_{b}}{{\log }_{b}}N}{{{\log }_{b}}a}}}$. We can write the given logarithmic expression as;
$\Rightarrow {{a}^{\dfrac{{{\log }_{b}}\left( {{\log }_{b}}N \right)}{{{\log }_{b}}a}}}$
We see that all the logarithms are present in the exponent of $a$. If we can express the logarithmic exponent without a fraction with the base $a$we can use the formula (1). So let us convert the base $b$ of the logarithm in the numerator to $d$ using the base change formula (2) for $x={{\log }_{b}}N$ We have;
$\Rightarrow {{a}^{\dfrac{\dfrac{{{\log }_{d}}\left( {{\log }_{b}}N \right)}{{{\log }_{d}}b}}{{{\log }_{b}}a}}}$
We change the base $b$ of the logarithm in the denominator for $x=a$ in the above step to have;
\[\Rightarrow {{a}^{\dfrac{\dfrac{{{\log }_{d}}\left( {{\log }_{b}}N \right)}{{{\log }_{d}}b}}{\dfrac{{{\log }_{d}}a}{{{\log }_{d}}b}}}}\]
We cancel out ${{\log }_{b}}d$ from the numerator and denominator to have;
\[\Rightarrow {{a}^{\dfrac{{{\log }_{d}}\left( {{\log }_{b}}N \right)}{{{\log }_{d}}a}}}\]
We use the base change formula (2) in reverse for $x={{\log }_{b}}N$ and base $b=a$ to change the base from $d$ to $a$ and have;
\[\Rightarrow {{a}^{{{\log }_{a}}\left( {{\log }_{b}}N \right)}}\]
We use formula (1) for $x={{\log }_{b}}N$ and $b=a$ to have;
\[\Rightarrow {{\log }_{b}}N\]

So, the correct answer is “Option A”.

Note: We note that in ${{\log }_{b}}x$ the value $x$ is called the argument of the logarithm and can only be positive. The base of the logarithm is always positive and is never equal to 1. The result of logarithm can be both positive, negative or zero. We must be careful of the common mistake of interchanging the numerator and denominator while using base change formula and must always write the logarithm of argument in the numerator. The logarithm with base 10 is called common logarithm which we can use instead of $d$ for easier understanding.