Question

If $\frac{{\log }_{a}N}{{\log }_{ab}N}=1+{\log }_{a}b$, then the permissible values of the variables are $\left[\begin{array}{cc}a>0;& a\ne 1\\ N>0;& N\ne 1\\ b>0;& b\ne \frac{1}{a}\end{array}\right]$.
If the above statement is true then write $1$ otherwise write $0$.

Answer 1

$\frac{{\log }_{a}N}{{\log }_{ab}N}=\frac{\frac{1}{{\log }_{N}a}}{\frac{1}{{\log }_{N}ab}}$

$=\frac{{\log }_{ab}N}{{\log }_{N}a}$
$=\frac{{\log }_{N}a+{\log }_{N}b}{{\log }_{N}a}$

$=1+\frac{{\log }_{N}b}{{\log }_{N}a}$

$\frac{{\log }_{a}N}{{\log }_{ab}N}=1+{\log }_{a}b$

Hence, the answer is $1$ as the statement is true.