If we have, say $$ \log_{10} (1/x)+\log_{10}(4/x)=-2, $$ then the solution is $$ x=\pm 20. $$ But the negative solution is false.
On the other hand, if we rewrite the equation to $$ \log_{10}(4/x^2)=-2, $$ then the solution is $$ x=\pm 20. $$
Negative solution is OK here since it is $x^2$ in the equation.
Even when I plot the equations it shows different solutions. So how is that working?
Rewriting $$\log_{10} (1/x)+\log_{10}(4/x)=-2\tag1$$ as $$\log_{10}(4/x^2)=-2\tag2$$ is to apply the identity $$\log_{10} (1/x)+\log_{10}(4/x)\equiv\log_{10}(4/x^2)$$ which has the implicit domain $x>0$ (due to its LHS); on $\mathbb R{\setminus}\{0\},$ equations $(1)$ and $(2)$ are not equivalent.
On the other hand, $$\log_{10}(4/x^2)=-2,\quad x>0\tag3$$ is an equivalent rewriting of equation $(1).$
In all simplicity, the input in the logarithm must be something positive. If some value of $x$ gives you a negative input, then you need to exclude it. In the first example, $x=-20$ gives $-\frac{1}{20}$ and $-\frac{1}{5}$ inside the logarithm, but in the second one it will be positive for both values of $x$.
The problem is that the functions you are using have defined values only for positive real numbers.
If you want to be able to take the log of a negative integer, you have to work with complex numbers, where ln(-1) = πi.
(It's easier to do with natural logs than with base 10.)
ln(-1/20) = πi + ln(1/20)
ln(-1/5) = πi + ln(1/5)
Sum = ln(1/20) + ln(1/5) + 2πi
When taking logs with complex numbers, 2nπi is equivalent to 0 (for all integers n), since e2nπi is zero.
So the sum of these two values will be the real number you're looking for.
