Question

Solve : $ {\log _x}\,2\, \times \,{\log _{\dfrac{x}{{16}}}}2\, = \,{\log _{\dfrac{x}{{64}}}}\,2 $

Answer 1

Hint: The logarithm is the inverse function of exponentiation in mathematics. That is, the logarithm of a given number \[{\text{x}}\] is the exponent to which another fixed number, the base $ {\text{b}} $ , must be increased in order to obtain the number $ {\text{x}} $ . The logarithm counts the number of times the same element appears in repeated multiplication in the simplest case.

Complete step-by-step answer:
Consider the equation $ {\log _x}\,2\, \times \,{\log _{\dfrac{x}{{16}}}}2\, = \,{\log _{\dfrac{x}{{64}}}}\,2 $
The change of base rule of logarithms can be used to swap the base and number in any logarithmic term.
Hence, by interchanging the numbers and base in log terms, we get
 $
   \Rightarrow \,\dfrac{1}{{{{\log }_2}x}}\, \times \,\dfrac{1}{{{{\log }_2}\left( {\dfrac{x}{{16}}} \right)}}\,\, = \,\,\dfrac{1}{{{{\log }_2}\left( {\dfrac{x}{{64}}} \right)}} \\
   \Rightarrow \,\dfrac{1}{{{{\log }_2}x\, \times \,{{\log }_2}\left( {\dfrac{x}{{16}}} \right)}}\,\, = \,\,\dfrac{1}{{{{\log }_2}\left( {\dfrac{x}{{64}}} \right)}} \;
  $
On further solving, using logarithmic rules, we get
 $
   \Rightarrow \,{\log _2}\left( {\dfrac{x}{{64}}} \right)\,\, = \,\,{\log _2}x\,\, \times \,{\log _2}\left( {\dfrac{x}{{16}}} \right) \\
   \Rightarrow \,{\log _2}x\, \times \,{\log _2}\left( {\dfrac{x}{{16}}} \right)\,\, = \,\,{\log _2}\left( {\dfrac{x}{{64}}} \right) \;
  $
There are two logarithmic terms that contain quotient-form quantities. The quotient rule of logarithmic can be used to expand them.
 $
   \Rightarrow {\log _2}x\, \times \,\left( {{{\log }_2}x\, - \,{{\log }_2}16} \right)\,\, = \,\,{\log _2}x\, - \,{\log _2}64 \\
   \Rightarrow {\left( {{{\log }_2}x} \right)^2}\, - \,\left( {{{\log }_2}{2^4}} \right)\,\, \times \,\left( {{{\log }_2}x} \right)\,\, = \,\,{\log _2}x\, - \,{\log _2}{2^6} \;
  $
To simplify the logarithmic equation, we use the power rule of logarithms.
 $ \Rightarrow {\left( {{{\log }_2}x} \right)^2}\, - \,\left( {4{{\log }_2}2} \right)\,\, \times \,\left( {{{\log }_2}x} \right)\,\, = \,\,{\log _2}x\, - \,6{\log _2}2 $
To simplify the equation even more, we use the logarithm of base law.
 $
   \Rightarrow \,{\left( {{{\log }_2}x} \right)^2}\, - \,\left( {4\, \times 1} \right)\, \times \,\left( {{{\log }_2}x} \right)\, = \,{\log _2}x\, - \,6 \times 1 \\
   \Rightarrow \,{\left( {{{\log }_2}x} \right)^2}\, - \,4{\log _2}x\, = \,{\log _2}x\, - \,6 \;
  $
On further solving we get,
 $
   \Rightarrow \,{\left( {{{\log }_2}x} \right)^2}\, - \,4{\log _2}x\, - \,{\log _2}x\, + 6\, = \,0 \\
   \Rightarrow \,{\left( {{{\log }_2}x} \right)^2}\, - 5\,{\log _2}x\, + 6\, = \,0 \;
  $
The above equation represents a quadratic equation. Hence, let us solve the quadratic equations and find the values.
Now, let us consider $ y\, = \,{\log _2}x $ , from logarithmic form, rewrite the equation in terms of $ y $ .
 $
   \Rightarrow \,{y^2}\, - 5y\, + \,6\, = \,0 \\
   \Rightarrow \,{y^2}\, - 3y\,\, - \,2y + \,6\, = \,0 \;
  $
On further solving we get,
 $ \Rightarrow \,y - 3\, = \,0\, $ and $ y\, - \,2\, = \,0 $
 $ \therefore \,\,y\, = \,3 $ and $ y\, = \,2 $
Hence, the values of $ y $ are $ 2 $ and $ 3 $ .
Replacing, the value for $ y $ , we get
 $ \therefore \,\,{\log _2}x\, = \,2 $ and $ {\log _2}x\, = \,3 $
To get the value of $ x $ mathematically, we write each logarithmic equation in exponential notation.
 $ \Rightarrow \,x\, = \,{2^2} $ and $ x\, = \,{2^3} $
 $ \therefore \,\,\,x\, = \,4 $ and $ x\, = \,8 $
So, the correct answer is “ $ \therefore \,\,\,x\, = \,4 $ and $ x\, = \,8 $ ”.

Note: John Napier invented logarithms in 1614 as a way of simplifying calculations. Navigators, scientists, engineers, surveyors, and others quickly embraced them to make high-accuracy computations easier. Logarithm tables can be used to replace repetitive multi-digit multiplication steps with table look-ups and easier addition.