Question

How does one solve $\log ({x^2}) = {\left( {\log x} \right)^2}$ ?

Answer 1

For simplifying the original equation , firstly used logarithm property $\log {a^b} = b\log a$ then take base ten exponential of both sides of the equation, then apply the logarithm formula ${b^{{{\log }_b}a}} = a$ to simplify the equation .

Formula used:
We used logarithm properties i.e.
$\log {a^b} = b\log a$ ,
And
${b^{{{\log }_b}a}} = a$ , the logarithm function says $\log x$ is only defined when $x$ is greater than zero.

Complete solution step by step:
It is given that ,
$\log {x^2} = {(\log x)^2} \\\ or \\\ 2\log x = {(\log x)^2} \\\$
Now , simplify the equation , we will get ,
$\log x(2\log x - \log x) = 0 \\\ or \\\ \log x = 2......(1) \\\ and \\\ \log x = 0....(2) \\\$
Now , by assuming the base of the logarithm to be $10$ ,then take the base $10$ exponential of both sides of the equation, we will get the following result ,
For equation one ,
${10^{{{\log }_{10}}(x)}} = {10^2}$
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get ,
$x = 100$
For equation two ,
${10^{{{\log }_{10}}(x)}} = {10^0}$ ,
By applying the logarithm formula ${b^{{{\log }_b}a}} = a$ . we will get ,
$x = 1$
Now recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero.
Therefore, in our original equation $\log ({x^2}) = {\left( {\log x} \right)^2}$ ,
Here,
$(x) > 0$ ,
For $x = 100$ and $x = 1$ ,
$100 > 0$ and $1 > 0$
Therefore, we have our solutions i.e., $100$ and $1$ .

Note: The logarithm function says $\log x$ is only defined when $x$ is greater than zero. While defining logarithm function one should remember that the base of the log must be a positive real number and not equals to one . At the end we must recall that the logarithm function says $\log x$ is only defined when $x$is greater than zero. While performing logarithm properties we have to remember certain conditions , our end result must satisfy the domain of that logarithm .