Question

The value of (loga n) / (logab n) is given by:

Answer 1

I have attached answer. Hope it helps.

Answer 2

Answer-

$\frac{\log_a n}{\log_{ab} n}=1+\log_a b}$

Solution-

The given expression is,

$\frac{\log_a n}{\log_{ab} n}$

From base change formula we know that,

$\log_y x=\frac{\log_z x}{\log_z y}$

Applying the same, taking the common base as 'e'

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=\frac{\frac{\log_e n}{\log_e a}}{\frac{\log_e n}{\log_e ab}}$

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=\frac{\log_e n}{\log_e a}\times \frac{\log_e ab}{\log_e n}$

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=\frac{\log_e ab}{\log_e a}$

From properties of logarithms, we know that,

$\log xy=\log x+\log y$

So,

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=\frac{\log_e a+\log_e b}{\log_e a}$

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=\frac{\log_e a}{\log_e a}+\frac{\log_e b}{\log_e a}$

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=1+\frac{\log_e b}{\log_e a}$

Applying reverse of base change formula,

$\Rightarrow \frac{\log_a n}{\log_{ab} n}=1+\log_a b}$