Divide equation by logarithm
I ran onto this problem when solving logarithmic equations.
Task is solve equation in R:
$x^{\log x - 3} = \frac{x}{1000}$
I understand that the correct solution is using substitution:
$\log x^{\log{x} - 3} = \log \frac{x}{1000}$
$(\log {x} - 3) \cdot \log {x} = \log x - \log 1000$
$(\log {x} - 3) \cdot \log {x} = \log{x} - 3$
substitution $s = \log{x}$
$(s - 3) \cdot s = s - 3$
$s^2 - 4\cdot s + 3 = 0$
$(s - 3)\cdot(s -1)= 0$
$s_1 = 3$, $s_2 = 1$
return
1)
$\log{x} = 3$
$x = 1000$
2)
$\log{x} = 1$
$x = 10$
$K = \{10; 1000\}$
But for the first time I did this and I am not sure why it gives me only one of solutions.
$log (x^{log(x) - 3}) = log(\frac{x}{1000})$
$log(\frac{x}{1000}) * (log (x)) = log(\frac{x}{1000})$
Then I divided equation with $log(\frac{x}{1000})$
$log (x) = 1$
$log(x) = log(10^1)$
$x = 10$
$K = \{10\}$
One solution is missing obviously. I have no clue what I did wrong, but I think that's is something about that division by $log(\frac{x}{1000})$.
Can someone tell what is wrong with this second method?
$\endgroup$ 11 Answer
$\begingroup$If you have the following equation: $$x^2=x$$ and you divide by $x$ to get $x=1$, you are assuming tha $x\neq 0$, as you cannot divide by zero. Therefore, you "eliminate" the $x=0$ solution, in a sense. Similarly, in your problem, when you divide by $\log(x/1000)$, you eliminate the solution in which this quantity is zero. In fact, this quantity is zero precisely when $x=1000$, which is the solution you missed.
So, in the future, whenever you divide by some quantity, you should ask yourself if/when the quantity is zero.
$\endgroup$
