Question

How do you solve ${{\left( \ln x \right)}^{2}}=\ln \left( {{x}^{2}} \right)$?

Answer 1

In the given question we have been asked to find the value of ‘x’ and it is given that ${{\left( \ln x \right)}^{2}}=\ln \left( {{x}^{2}} \right)$. In order to solve the question, first we need to use the basic property of logarithms i.e. $\ln \left( {{a}^{b}} \right)=b\ln \left( x \right)$ and ${{\log }_{b}}\left( x \right)=y$ is equivalent to${{b}^{y}}=x$. Then we simplify the equation further to get the possible values of ‘x’.

Formula used:
$\ln \left( {{a}^{b}} \right)=b\ln \left( x \right)$
If $x$ and b are positive real numbers and b is not equal to 1,
Then ${{\log }_{b}}\left( x \right)=y$ is equivalent to ${{b}^{y}}=x$.

Complete step by step solution:
We have given that,
${{\left( \ln x \right)}^{2}}=\ln \left( {{x}^{2}} \right)$
As, we know that,
$\ln \left( {{a}^{b}} \right)=b\ln \left( x \right)$
Applying this in the given equation, we get
$\Rightarrow {{\left( \ln x \right)}^{2}}=2\ln \left( x \right)$
Substitute ln (x) = k,
Now, solving the equation, we get
$\Rightarrow {{k}^{2}}=2k$
Write the above equation in the standard form, we get
$\Rightarrow {{k}^{2}}-2k=0$
Taking out ‘k’ as a common factor, we get
$\Rightarrow k\times \left( k-2 \right)=0$
Solving each term individually, we get
$\Rightarrow k=0$ And $k-2=0$
$\Rightarrow k=0$ And $k=2$
Now, undo the substitution i.e. k = ln (x), we get
$\Rightarrow \ln \left( x \right)=0$ and $\ln \left( x \right)=2$
Now, solving
$\Rightarrow \ln \left( x \right)=0$
Using the definition of log,
If $x$ and b are positive real numbers and b is not equal to 1,
Then ${{\log }_{b}}\left( x \right)=y$is equivalent to${{b}^{y}}=x$.
$\Rightarrow {{e}^{0}}=x$
$\Rightarrow x=1$
Similarly, solving
$\Rightarrow \ln \left( x \right)=2$
$\Rightarrow {{e}^{2}}=x$
$\Rightarrow x={{e}^{2}}$
Therefore, the possible values of ‘x’ are 1 and ${{e}^{2}}$.
It is the required solution.

Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations.