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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?

Glorfindel
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devWeSp
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1 Answers1

4

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.

florence
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