Question

Solve the equation :

\log_{4} (\log_{2} x) + \log_{2} (\log_{4} x) = 2

Accepted Answer

It is given $\log_{4} (\log_{2} x) + \log_{2} (\log_{4} x) = 2$

We know that $\log_{a^{n}} b = \frac{1}{n} \log_{a} b, b > 0, a > 0, a \neq 1$ , using this, we can write,

$\Rightarrow \log_{2^{2}} (\log_{2} x) + \log_{2} (\log_{4} x) = 2$

$\Rightarrow \frac{1}{2} \log_{2} (\log_{2} x) + \log_{2} (\log_{4} x) = 2$

$\Rightarrow \log_{2} (\log_{2} x) + 2 \log_{2} (\log_{4} x) = 4$

We know that $\log_{a} b^{m} = m \log_{a} b, b > 0, a > 0, a \neq 1$ , using this, we can write,

$\Rightarrow \log_{2} (\log_{2} x) + \log_{2} {(\log_{4} x)}^{2} = 4$

We know that $\log_{a} m + \log_{a} n = \log_{a} (m n), m > 0, n > 0, a > 0, a \neq 1$ , using this, $\log_{2} (\log_{2} x) + \log_{2} {(\log_{4} x)}^{2} = 4$ can be written as,

$\Rightarrow \log_{2} \{(\log_{2} x) {(\log_{4} x)}^{2}\} = 4$

Using $\log_{a} b = N \Leftrightarrow b = a^{N}, b > 0, a > 0, a \neq 1$ , we can write,

$\Rightarrow \{(\log_{2} x) {(\log_{4} x)}^{2}\} = 2^{4}$

$\Rightarrow \{(\log_{2} x) {(\frac{\log_{2} x}{2})}^{2}\} = 16$

$\Rightarrow \{{(\log_{2} x)}^{3}\} = 64$

$\Rightarrow \log_{2} x = 4$

Using the property of logarithm, $\log_{a} b = N \Leftrightarrow b = a^{N}, b > 0, a > 0, a \neq 1$ , we can write,

$\Rightarrow x = 2^{4}$

$\Rightarrow x = 16$ .

Embibe Experts Solutions for Chapter: Logarithm and its Applications, Exercise 1: Exercise-1

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Logarithm and its Applications, Exercise 1: Exercise-1

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