QUESTION

# Solve the following equation:${\log _{16}}x + {\log _4}x + {\log _2}x = 7$

Hint: We have to use the necessary logarithmic properties to find the value of x.

It is given to us that ${\log _{16}}x + {\log _4}x + {\log _2}x = 7$
We know that $\left[ {\because {{\log }_a}b = \dfrac{1}{{{{\log }_b}a}}} \right]$
So, we will get
$\Rightarrow \dfrac{1}{{{{\log }_x}16}} + \dfrac{1}{{{{\log }_x}4}} + \dfrac{1}{{{{\log }_x}2}} = 7$
On simplification, we get
$\Rightarrow \dfrac{1}{{{{\log }_x}{2^4}}} + \dfrac{1}{{{{\log }_x}{2^2}}} + \dfrac{1}{{{{\log }_x}2}} = 7$
Now since we know that $\left[ {\because n{{\log }_a}M = {{\log }_a}{M^n}} \right]$
And hence on following the above formula we have,
$\Rightarrow \dfrac{1}{{4{{\log }_x}2}} + \dfrac{1}{{2{{\log }_x}2}} + \dfrac{1}{{{{\log }_x}2}} = 7$
And hence on taking $\dfrac{1}{{{{\log }_x}2}}$ common, we get,
$\left[ {\dfrac{1}{4} + \dfrac{1}{2} + 1} \right]\dfrac{1}{{{{\log }_x}2}} = 7$
And hence on doing the simplification, we have
$\left[ {\dfrac{7}{4}} \right]\dfrac{1}{{{{\log }_x}2}} = 7$
$\Rightarrow \dfrac{1}{{{{\log }_x}2}} = 7 \times \left[ {\dfrac{4}{7}} \right]$
Here 7 will get cancelled out
$\Rightarrow {\log _2}x = 4$ $\left[ {\because {{\log }_a}b = \dfrac{1}{{{{\log }_b}a}}} \right]$
$\Rightarrow {2^4} = x$
$\Rightarrow x = 16$

Note: This question consists of equations comprising logarithmic functions. So we just need to use the appropriate logarithmic properties. Mistakes should be avoided in application of these properties.