Question

If $x > 0$ and ${\log _3}x + {\log _3}\left( {\sqrt x } \right) + {\log _3}\left( {\sqrt[4]{x}} \right) + ... = 4$ , then $x$ equals
(1) $9$
(2) $81$
(3) $1$
(4) $27$

Answer 1

Hint: The logarithm function is the inverse function of exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. We use a general coefficient rule for this problem i.e. $\log {x^m} = m\log x$ where $x > 0$ and m is a real number.

Complete step by step answer:
First we collect the data from the given question and analysis it
Given ${\log _3}x + {\log _3}\left( {\sqrt x } \right) + {\log _3}\left( {\sqrt[4]{x}} \right) + ... = 4$ where $x > 0$
We know the formula $\sqrt[n]{a} = {a^{\dfrac{1}{n}}}$ , use this formula in above equation and we get
${\log _3}x + {\log _3}\left( {x^{\dfrac{1}{2}} } \right) + {\log _3}\left( {x^{\dfrac{1}{4}}} \right) + ... = 4$ where $x > 0$
Now using the formula $\log {x^m} = m\log x$where $x > 0$ and m is real number,
we get ${\log _3}x + \dfrac{1}{2}{\log _3}x + \dfrac{1}{4}{\log _3}x + ... = 4$
Take common ${\log _3}x$ from the above equation we get
i.e. ${\log _3}x[1 + \dfrac{1}{2} + \dfrac{1}{4} + ...] = 4$
Applying property of geometric progression and we get
i.e. ${\log _3}x\left( {\dfrac{1}{{1 - \dfrac{1}{2}}}} \right) = 4$
i.e., ${\log _3}x\left( {\dfrac{1}{{\dfrac{1}{2}}}} \right) = 4$
i.e., ${\log _3}x\left( 2 \right) = 4$
dividing $2$ from both sides of the above equation and we get
i.e. ${\log _3}x = 2$
Use the property of logarithm and we get
i.e. $x = {3^2}$
i.e., $x = 9$

So, the correct answer is “Option 1”.

Note:
Note that the series $1 + \dfrac{1}{2} + \dfrac{1}{4} + ...$ is called a geometric series with infinite terms. The method to find its summation is $\dfrac{a}{{1 - r}}$ where a=the first term of the series and r=common difference between the terms of the series.
Hence
$1 + \dfrac{1}{2} + \dfrac{1}{4} + ...=\dfrac{1}{{1 - \dfrac{1}{2} }}=2$
Also we know that if ${\log _p}c = b$ then ${p^b} = c$ . Thus ${\log _3}x = 2$ implies that $x = {3^2} = 9$.