Question

How do you solve ${\log _2}x + {\log _4}x + {\log _8}x + {\log _{16}}x = 25$ ?

Answer 1

Hint: This given question is related to logarithm. In order to solve this question, the concepts of logarithm and all the related logarithm properties should be clear. An exponent that is written in a special way is known as a logarithm. Logarithm functions are just opposite or inverse of exponential functions. We can easily express any exponential function in a logarithm form. Similarly, all the logarithm functions can be easily rewritten in exponential form. Here in this question, we are given a logarithm equation and we will have to use the necessary logarithm properties to find the value of $x$.

Complete step by step answer:
Given is ${\log _2}x + {\log _4}x + {\log _8}x + {\log _{16}}x = 25$
We know the following logarithm property:
${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$
Using the above logarithm property in the given logarithm equation, we get,
$ \Rightarrow \dfrac{1}{{{{\log }_x}2}} + \dfrac{1}{{{{\log }_x}4}} + \dfrac{1}{{{{\log }_x}8}} + \dfrac{1}{{{{\log }_x}16}} = 25$
On simplification we get,
$ \Rightarrow \dfrac{1}{{{{\log }_x}2}} + \dfrac{1}{{{{\log }_x}{2^2}}} + \dfrac{1}{{{{\log }_x}{2^3}}} + \dfrac{1}{{{{\log }_x}{2^4}}} = 25$-----(1)
We also know that:$\left[ {\because n{{\log }_a}M = {{\log }_a}{M^n}} \right]$
Using the above property in equation (1)
$ \Rightarrow \dfrac{1}{{{{\log }_x}2}} + \dfrac{1}{{2{{\log }_x}2}} + \dfrac{1}{{3{{\log }_x}2}} + \dfrac{1}{{4{{\log }_x}2}} = 25$
Taking$\dfrac{1}{{{{\log }_x}2}}$common from the above equation, we get,
$\dfrac{1}{{{{\log }_x}2}}\left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}} \right) = 25 \\
\Rightarrow {\log _x}2\left( {\dfrac{{12 + 6 + 4 + 3}}{{12}}} \right) = 25 \\
\Rightarrow {\log _x}2\left( {\dfrac{{25}}{{12}}} \right) = 25 \\
\Rightarrow {\log _x}2 = 25 \times \dfrac{{12}}{{25}} \\
\Rightarrow {\log _x}2 = 12 \\ $
Now again using the logarithm property, ${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$, we get,
$x = {2^{12}} \\
\therefore x = 4096 \\ $
Therefore, the value of $x$ is $4096$.

Note: Here in this question, the equation consists of logarithmic functions. In order to solve this question without any complexity, the students must be aware of the logarithm functions concept as well as various logarithm properties. The most common mistakes while doing such types of questions are calculation mistakes. Students should carefully do the calculations and try to avoid any mistake in application of logarithm properties. Logarithms are useful when we want to work with large numbers. Logarithm has many uses in real life, such as in electronics, acoustics, earthquake analysis and population prediction.