Question

The solution of ${{\log }_{\sqrt{3}}}x+{{\log }_{\sqrt[4]{3}}}x+{{\log }_{\sqrt[6]{3}}}x+.....+{{\log }_{\sqrt[16]{3}}}x=36$ is?
(a) $x=3$
(b) $x=4\sqrt{3}$
(c) $x=9$
(d) $x=\sqrt{3}$

Answer 1

Use the formula of exponents of radical terms given as $\sqrt[p]{x}={{\left( x \right)}^{\dfrac{1}{p}}}$ to simplify the bases of the logarithmic terms. Now, use the formula of logarithms ${{\log }_{{{a}^{m}}}}b=\dfrac{1}{m}{{\log }_{a}}b$ and take $\log x$ common from all the terms. Use the formula of sum of first n even natural numbers equal to $n\left( n+1 \right)$ and cancel the common factors from both the sides. Finally, convert the log function into the exponential function by using the relation conversion given as ‘if ${{\log }_{a}}x=k$ then $x={{a}^{k}}$’ to get the answer.

Complete step-by-step solution:
Here we have been provided with the expression ${{\log }_{\sqrt{3}}}x+{{\log }_{\sqrt[4]{3}}}x+{{\log }_{\sqrt[6]{3}}}x+.....+{{\log }_{\sqrt[16]{3}}}x=36$ and we are asked to find the solution that means the value of x for the equation.
Now, we know that the radical expression $\sqrt[p]{x}$ can be converted into the exponential form by using the relation $\sqrt[p]{x}={{\left( x \right)}^{\dfrac{1}{p}}}$, so simplifying the bases of the logarithmic terms we get,
$\Rightarrow {{\log }_{{{\left( 3 \right)}^{\dfrac{1}{2}}}}}x+{{\log }_{{{\left( 3 \right)}^{\dfrac{1}{4}}}}}x+{{\log }_{{{\left( 3 \right)}^{\dfrac{1}{6}}}}}x+.....+{{\log }_{{{\left( 3 \right)}^{\dfrac{1}{16}}}}}x=36$
Using the property of logarithm given as ${{\log }_{{{a}^{m}}}}b=\dfrac{1}{m}{{\log }_{a}}b$ we get,

& \Rightarrow \dfrac{1}{\left( \dfrac{1}{2} \right)}{{\log }_{3}}x+\dfrac{1}{\left( \dfrac{1}{4} \right)}{{\log }_{3}}x+\dfrac{1}{\left( \dfrac{1}{6} \right)}{{\log }_{3}}x+.....+\dfrac{1}{\left( \dfrac{1}{16} \right)}{{\log }_{3}}x=36 \\\ & \Rightarrow 2{{\log }_{3}}x+4{{\log }_{3}}x+6{{\log }_{3}}x+.....+16{{\log }_{3}}x=36 \\\ & \Rightarrow \left( 2+4+6+.....+16 \right){{\log }_{3}}x=36 \\\ \end{aligned}$$ We know that the sum of first n even natural numbers is equal to $n\left( n+1 \right)$, so in the above relation we have 8 terms counting from 2 to 16 inside the bracket, therefore we get, $$\begin{aligned} & \Rightarrow \left( 8\times 9 \right){{\log }_{3}}x=36 \\\ & \Rightarrow 72{{\log }_{3}}x=36 \\\ & \Rightarrow {{\log }_{3}}x=\dfrac{1}{2} \\\ \end{aligned}$$ The inter-conversion relation of the log and the exponent is given as ‘if ${{\log }_{a}}x=k$ then $x={{a}^{k}}$’, so we get, $$\begin{aligned} & \Rightarrow x={{\left( 3 \right)}^{\dfrac{1}{2}}} \\\ & \therefore x=\sqrt{3} \\\ \end{aligned}$$ **Hence, option (d) is the correct answer.** **Note:** Note that logarithm is the inverse operation of exponentiation. For a logarithmic term ${{\log }_{a}}b$ to be defined we must have its argument (b) and the base (a) greater than 0 and also the base value must not be equal to 1. In case we would have obtained two solutions for the value of x then we would have rejected the negative value as it makes the function undefined.