Question

Find the value of \[x\] if \[{{\log }_{10}}\left( {{x}^{2}}-12x+36 \right)=2\].

Answer 1

Hint: The logarithmic and exponential functions are opposite to each other. i.e. \[{{\log }_{e}}\left( {{e}^{x}} \right)=x\].In this case, we have logarithm to the base 10, so we will convert the logarithm to base \[e\] and take exponential of both sides.

Formula used:
In order to convert natural log to logarithm to base 10, we multiply the natural logarithmic term with 2.303.
\[{{\log }_{10}}x=\ln x\times 2.303\]
Secondly, the exponential and natural logarithm are opposite to each other, so they cancel out, i.e. natural log of exponential term of a number or vice versa is the number itself.
\[{{\log }_{e}}\left( {{e}^{x}} \right)=x\]
\[{{e}^{{{\log }_{e}}x}}=x\]

Complete step by step answer:
Writing the equation and converting the logarithm to the base\[e\].
\[{{\log }_{10}}\left( {{x}^{2}}-12x+36 \right)=2\]
Now, converting the logarithm form to natural log
\[\begin{align}
  & \dfrac{{{\log }_{e}}\left( {{x}^{2}}-12x+36 \right)}{2.303}=2 \\
 & {{\log }_{e}}\left( {{x}^{2}}-12x+36 \right)=2\times 2.303 \\
 & {{\log }_{e}}\left( {{x}^{2}}-12x+36 \right)=4.606 \\
\end{align}\]
Now, taking exponential of both sides, where exponential and logarithm terms cancel each other as they are of opposite nature.
\[\begin{align}
  & {{e}^{\left\{ {{\log }_{e}}\left( {{x}^{2}}-12x+36 \right) \right\}}}={{e}^{4.606}} \\
 & \Rightarrow {{x}^{2}}-12x+36=100.08 \\
 & \Rightarrow {{x}^{2}}-12x+36\simeq 100 \\
 & \Rightarrow {{x}^{2}}-12x+36-100=0 \\
 & \Rightarrow {{x}^{2}}-12x+64=0 \\
 & \Rightarrow {{x}^{2}}-16x+4x-64=0 \\
 & \Rightarrow x\left( x-16 \right)+4\left( x-16 \right)=0 \\
 & \Rightarrow \left( x-16 \right)\left( x+4 \right)=0 \\
 & \Rightarrow x=16,-4 \\
\end{align}\]

Additional information:
A logarithm is the power to which a number must be raised in order to get some other number. For example, the base ten logarithm of \[100\] is \[2\], because ten raised to the power of two is \[100\] : \[{{\log }_{10}}100=2\] because\[{{10}^{2}}=100\]. Whereas the exponential is just the opposite , it is of the form \[f\left( x \right)={{a}^{x}}\].

Note: Alternatively, this can also be solved without converting the log to the base 10 to the natural logarithm. In this case we take the \[{{10}^{x}}\] exponential of the both sides and then solve. In this case, the right hand side becomes \[{{10}^{2}}\] and the left hand side becomes \[{{x}^{2}}-12x+36\].
\[\begin{align}
  & {{\log }_{10}}\left( {{x}^{2}}-12x+36 \right)=2 \\
 & {{x}^{2}}-12x+36=100 \\
 & {{x}^{2}}-12x-64=0 \\
 & {{x}^{2}}-16x+4x-64=0 \\
 & x\left( x-16 \right)+4\left( x-16 \right)=0 \\
 & \left( x-16 \right)\left( x+4 \right)=0 \\
 & x=16,-4 \\
\end{align}\]
Thus, the values come out to be the same for this method also.