Solve the equation,$\sqrt{\log(-x)}=\log{\sqrt{x^2}}$(base of log is 10)
$\sqrt{\log(-x)}=\log{\sqrt{x^2}}$
$\sqrt{\log(-x)}=\log{|x|}$
Now two cases arise,when $x>0$ and when $x<0$
When $x<0$,
$\sqrt{\log(-x)}=\log(-x)$
I found $x=-1,-10$
When $x>0$
$\sqrt{\log(-x)}=\log x$
$\log(-x)=\log x\times\log x$
I could not solve it further.Please help me.
On
Since the base of exponential is 10, the correct value for $x$ will be $-1$ and $-10$.In other words, if $\sqrt{\log(-x)}=\log{-x}$ then the equation has solution. Check by putting in the value.
On
It is a godo approach to split into cases depending on the sign of $x$. Clearly $x=0$ is not a solution, so you just consider (as you do) $x >0$ and $x<0$.
If $x >0$, then $\log(-x)$ isn't defined because $-x<0$, so there are not solutions where $x > 0$.
For $x<0$ you find (as you do) $\sqrt{\log(-x)} = \log(-x)$. The solution to the equation $a^2 = a$ is $a =0$ or $a = 1$. So you have two possibilities $$ \log(-x) = 0 \quad\text{or}\quad \log(-x) = 1. $$ If $\log(-x) = 0$, then $-x = 1$, so $x = -1$. If $\log(-x) = 1$, then $-x = 10$, and so $x=-10$.
In all you have exactly these two solutions and no other.
$\log(−x)$ for $x>0$ would be a logarithm of negative number, and it doesn't exist.