Question

How do you solve \[\log \left( 2x \right)-\log \left( {{2}^{x}} \right)=\log \left( {{x}^{2}} \right)-\log \left( x \right)\] ?

Answer 1

Hint: The above equation can be done by remembering the basic form and formulae of logarithms. The basic form of logarithm is expressed as:
\[{{\log }_{b}}{{a}^{x}}=x{{\log }_{b}}a\] Here \[b\] is the base, \[a\] is the exponent and \[x\] is the power.
Now, if \[a=b\] , \[{{\log }_{a}}{{a}^{x}}=x{{\log }_{a}}a=x\] as we know \[{{\log }_{a}}a=1\] .
The basic formulae for logarithms are:
\[\begin{align}
  & \log a+\log b=\log \left( ab \right) \\
 & \log a-\log b=\log \left( \dfrac{a}{b} \right) \\
\end{align}\]

Complete step by step answer:
When no base has been mentioned in the problem, we assume the base to be as \[10\] . We also need to remember that, if \[{{\log }_{10}}a=c\] then we can easily find out the value of \[a\] as, \[a={{10}^{c}}\].
Applying one basic formula in the question, which states that,
\[\log a-\log b=\log \left( \dfrac{a}{b} \right)\]
We can write,
\[\log \left( 2x \right)-\log \left( {{2}^{x}} \right)=\log \left( {{x}^{2}} \right)-\log \left( x \right)\]
Now, using the logarithmic rules in the above equation on both the L.H.S and R.H.S we get,
\[\log \left( \dfrac{2x}{{{2}^{x}}} \right)=\log \left( \dfrac{{{x}^{2}}}{x} \right)\]
Now, taking antilog on both sides, we get the following equation,
\[\dfrac{2x}{{{2}^{x}}}=\dfrac{{{x}^{2}}}{x}\]
Now, cross multiplying both sides we get,
\[\begin{align}
  & 2{{x}^{2}}={{x}^{2}}{{2}^{x}} \\
 & \Rightarrow 2{{x}^{2}}-{{x}^{2}}{{2}^{x}}=0 \\
\end{align}\]
Now, taking \[{{x}^{2}}\] common, we get,
\[{{x}^{2}}\left( 2-{{2}^{x}} \right)=0\]
Now, for this above equation, we have two solutions. The first solution may be written as,
\[\begin{align}
  & {{x}^{2}}=0 \\
 & \Rightarrow x=0 \\
\end{align}\]
But, this solution of \[x\] is not a valid one because, if we put \[x=0\] in our original given equation, \[\log 0\] becomes undefined.
The other solution may be written as,
\[\begin{align}
  & \left( 2-{{2}^{x}} \right)=0 \\
 & \Rightarrow 2={{2}^{x}} \\
 & \Rightarrow {{2}^{1}}={{2}^{x}} \\
 & \Rightarrow x=1 \\
\end{align}\]
This is a valid solution, as if we put \[x=1\] in our original given equation, it satisfies both the L.H.S and R.H.S.

So, the correct answer is \[x=1\].

Note: We need to keep in mind of all the logarithmic formulae and relations for smooth solving of these types of questions. We also have to keep in mind that \[\log 0\] is undefined and hence to avoid such kinds of complications in our solution.