Question

Simplify $ {{\log }_{7}}{{\log }_{7}}\sqrt{7\sqrt{7\sqrt{7}}} $ \[\]
A. $ 3{{\log }_{2}}7 $ \[\]
B. $ 1-3{{\log }_{3}}7 $ \[\]
C. $ 1-3{{\log }_{7}}2 $ \[\]
D. None of these \[\]

Answer 1

Hint: We start by simplifying the innermost square root in the given expression using the exponential identities like $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ and $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $ until the all three square roots vanish. We simplify the logarithm using identities of power $ m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}} $ and quotient $ {{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n $ .

Complete step-by-step answer:
We know that the logarithm is the inverse operation to 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 $ , which means if $ {{b}^{y}}=x $ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
Here $ x,y,b $ are real numbers subjected to condition $ x > 0 $ and $ b > 0,b\ne 1 $
We know that
\[{{\log }_{b}}b=1\]
We know the logarithmic identity involving power $ m\ne 0 $ where $ m $ is real number as
\[m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}\]
We also know the logarithmic identity involving quotient as
\[{{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n\]

We are given in the question the numerical expression involving logarithm as
\[{{\log }_{7}}{{\log }_{7}}\sqrt{7\sqrt{7\sqrt{7}}}\]
We begin with simplifying the innermost square root. We know that $ \sqrt{a}={{\left( a \right)}^{\dfrac{1}{2}}} $ for any real number $ a\ne 0 $ . So we have,
\[{{\log }_{7}}{{\log }_{7}}\sqrt{7\sqrt{7\sqrt{7}}}={{\log }_{7}}{{\log }_{7}}\sqrt{7\sqrt{7\times {{7}^{\dfrac{1}{2}}}}}\]
We use the exponential identity $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $ where $ n $ is real number in above step for $ a=7,m=1,n=\dfrac{1}{2} $ to have,
\[\begin{align}
  & ={{\log }_{7}}{{\log }_{7}}\sqrt{7\sqrt{{{7}^{1+\dfrac{1}{2}}}}} \\
 & ={{\log }_{7}}{{\log }_{7}}\sqrt{7\sqrt{{{7}^{\dfrac{3}{2}}}}} \\
 & ={{\log }_{7}}{{\log }_{7}}\sqrt{7{{\left( {{7}^{\dfrac{3}{2}}} \right)}^{\dfrac{1}{2}}}} \\
\end{align}\]
We use the exponential identity $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ where $ {{a}^{m}} $ is a nonzero real number in above step for $ a=7,m=\dfrac{3}{2},n=\dfrac{1}{2} $ to have,
\[\begin{align}
  & ={{\log }_{7}}{{\log }_{7}}\sqrt{7\times {{7}^{\dfrac{3}{2}\times \dfrac{1}{2}}}} \\
 & ={{\log }_{7}}{{\log }_{7}}\sqrt{7\times {{7}^{\dfrac{3}{4}}}} \\
\end{align}\]
We use the exponential identity $ {{a}^{m}}\times {{a}^{n}}={{a}^{m+n}} $ for $ a=7,m=1,n=\dfrac{3}{4} $ in above step to have,
\[\begin{align}
  & ={{\log }_{{}}}{{\log }_{7}}\sqrt{{{7}^{1+\dfrac{3}{4}}}} \\
 & ={{\log }_{7}}{{\log }_{7}}\sqrt{{{7}^{\dfrac{7}{4}}}} \\
 & ={{\log }_{7}}{{\log }_{7}}{{\left( {{7}^{\dfrac{7}{4}}} \right)}^{\dfrac{1}{2}}} \\
\end{align}\]
We use the exponential identity $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}} $ for $ a=7,m=\dfrac{7}{4},n=\dfrac{1}{2} $ in above step to have,
\[\begin{align}
  & ={{\log }_{7}}{{\log }_{7}}{{7}^{\dfrac{7}{4}\times \dfrac{1}{2}}} \\
 & ={{\log }_{7}}{{\log }_{7}}{{7}^{\dfrac{7}{8}}} \\
\end{align}\]
We use the logarithmic identity of quotient $ m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}} $ for $ b=x=7,m=\dfrac{7}{8} $ in above step to have,
\[={{\log }_{7}}\dfrac{7}{8}{{\log }_{7}}7\]
We use the logarithmic identity $ {{\log }_{b}}b=1 $ for $ b=7 $ in above step to have,
\[={{\log }_{7}}\dfrac{7}{8}\]
We use the logarithmic identity of quotient $ {{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n $ for $ b=7,m=7,n=8 $ in above step to have,
\[\begin{align}
  & ={{\log }_{7}}7-{{\log }_{7}}8 \\
 & =1-{{\log }_{7}}8 \\
\end{align}\]
We find the find the prime factorization of 8 and replace 8 as $ 8=2\times 2\times 2={{2}^{3}} $ in above step to have,
\[=1-{{\log }_{7}}{{2}^{3}}\]
We use the logarithmic identity of quotient $ m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}} $ for $ b=7,x=2,m=3 $ in above step to have,
\[=1-3{{\log }_{7}}2\]

So, the correct answer is “Option C”.

Note: The logarithmic identity of product is given by $ {{\log }_{b}}\left( mn \right)={{\log }_{b}}m+{{\log }_{b}}n $ . The logarithm is called binary logarithm when the base is 2 and denoted as $ \operatorname{lb}x $ , natural logarithm when the base is transcendental $ e $ and denoted as $ \ln x $ and common logarithm when the base is 10 and denoted as $ \lg x. $