Question

Solve the following expression:
 \[2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}2\sqrt{2} \right)=1\].

Answer 1

Hint: In this question, we need to convert the square root terms inside the logarithm accordingly using the identities to simplify it. Then rearrange the terms and do basic operations like taking the terms outside the logarithm to inside and solve further to get the value of x.

Complete step by step answer:
\[\begin{align}
  & {{a}^{{{\log }_{a}}x}}=x,a\ne 0,\ne \pm 1,x>0 \\
 & {{\log }_{a}}a=1,{{\log }_{a}}1=0;a>0,\ne 1 \\
\end{align}\]
For m, n > 0, a > 0 and a not equal to 1
\[{{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\]
For x > 0, a > 0 and a not equal to 1
\[\begin{align}
  & {{\log }_{{{a}^{n}}}}\left( x \right)=\dfrac{1}{n}{{\log }_{a}}x \\
 & {{\log }_{{{a}^{n}}}}{{x}^{m}}=\dfrac{m}{n}{{\log }_{a}}x \\
\end{align}\]

Let us look at some of the basic definitions of logarithm.
If a is a positive real number other than 1 and \[{{a}^{x}}=m\] , then x is called the logarithm of m to the base a, written as \[{{\log }_{a}}m\] . In \[{{\log }_{a}}m\] , m should always be positive.
If m < 0, then \[{{\log }_{a}}m\] will be imaginary and if m = 0, then \[{{\log }_{a}}m\] will be meaningless.
\[{{\log }_{a}}m\] exists only, if m, a > 0 and a is not equal to 1.
TYPES OF LOGARITHMS:
Natural or Napier Logarithms: The logarithm with base 'e' (e = 2.718) is called natural logarithm.
Common Logarithms: The logarithm with base 10 is called common logarithm.
CHARACTERISTIC AND MANTISSA OF A LOGARITHM:
The logarithm of positive real number n consists of two parts:
The integral part is known as the characteristic. It is always an integer positive, negative or zero.
The decimal part is called the mantissa . The mantissa is never negative and is always less than one.
ANTI LOGARITHM: The positive number a is called the antilogarithm of a number b, if a is anti logarithm of b, then we write a = antilog b.
So, \[a=anti\log b\Leftrightarrow \log a=b\]
Let us list few laws of the logarithm.
\[\begin{align}
  & {{a}^{{{\log }_{a}}x}}=x,a\ne 0,\ne \pm 1,x>0 \\
 & {{\log }_{a}}a=1,{{\log }_{a}}1=0;a>0,\ne 1 \\
\end{align}\]
For m, n > 0, a > 0 and a not equal to 1
\[{{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n\]
For x > 0, a > 0 and a not equal to 1
\[\begin{align}
  & {{\log }_{{{a}^{n}}}}\left( x \right)=\dfrac{1}{n}{{\log }_{a}}x \\
 & {{\log }_{{{a}^{n}}}}{{x}^{m}}=\dfrac{m}{n}{{\log }_{a}}x \\
\end{align}\]
Now, from the equation given in the question we can write,
\[\begin{align}
  & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}2\sqrt{2} \right)=1 \\
 & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{\dfrac{1}{2}}}\left( {{\log }_{2}}{{2}^{\dfrac{3}{2}}} \right)=1 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{\dfrac{1}{2}}}\dfrac{3}{2}\left( {{\log }_{2}}2 \right)=1 \\
 & \left[ \because {{\log }_{{{a}^{n}}}}{{x}^{m}}=\dfrac{m}{n}{{\log }_{a}}x \right] \\
\end{align}\]
Now, on further simplification we get,
\[\begin{align}
  & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{\dfrac{1}{2}}}\dfrac{3}{2}=1 \\
 & \left[ \because {{\log }_{a}}a=1 \right] \\
\end{align}\]
\[\Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{\dfrac{1}{2}}}3-{{\log }_{\dfrac{1}{2}}}2=1\]
\[\left[ \because {{\log }_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n \right]\]
\[\begin{align}
  & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)+{{\log }_{{{2}^{-1}}}}3-{{\log }_{{{2}^{-1}}}}2=1 \\
 & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)-{{\log }_{2}}3+{{\log }_{2}}2=1 \\
\end{align}\]
\[\left[ \because {{\log }_{{{a}^{n}}}}\left( x \right)=\dfrac{1}{n}{{\log }_{a}}x \right]\]
\[\begin{align}
  & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)-{{\log }_{2}}3=1-1 \\
 & \left[ \because {{\log }_{a}}a=1 \right] \\
 & \Rightarrow 2{{\log }_{2}}\left( {{\log }_{2}}x \right)={{\log }_{2}}3 \\
 & \Rightarrow {{\log }_{2}}\left( {{\log }_{2}}x \right)=\dfrac{1}{2}{{\log }_{2}}3 \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{\log }_{2}}\left( {{\log }_{2}}x \right)={{\log }_{2}}{{3}^{\dfrac{1}{2}}} \\
 & \Rightarrow {{\log }_{2}}x=\sqrt{3} \\
 & \therefore x={{2}^{\sqrt{3}}}\text{ }\left[ \because {{a}^{x}}=m\Leftrightarrow x={{\log }_{a}}m \right] \\
\end{align}\]

Note: A negative number can never be expressed as the power of 10, mantissa should always be kept positive. Hence, whenever a characteristic is negative, a minus sign is placed above the characteristic and not to its left to show that the mantissa is always positive.
While applying the laws of logarithm we need to be careful with the sign and also the value of the coefficient when we bring it outside the logarithm because it may change the base or index of any other term so that we could not solve it further.
Instead of writing it in terms of subtraction because of considering it as a division term we can also consider it as a multiplication term and then write it as summation which also gives the same result.
\[\begin{align}
  & \Rightarrow {{\log }_{\dfrac{1}{2}}}\dfrac{3}{2} \\
 & \left[ \because {{\log }_{a}}\left( mn \right)={{\log }_{a}}m+{{\log }_{a}}n \right] \\
\end{align}\]
\[\Rightarrow {{\log }_{\dfrac{1}{2}}}3+{{\log }_{\dfrac{1}{2}}}\dfrac{1}{2}\]
\[\begin{align}
  & \Rightarrow {{\log }_{{{2}^{-1}}}}3+1 \\
 & \left[ \because {{\log }_{a}}a=1 \right] \\
\end{align}\]