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I may be being stupid right now, so I've come to Stack to see if this elementary algebra holds up.

Suppose I have the equation $$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$$

My chosen way to solve this would be to cross multiply and expand brackets, solve the quadratic and get the value of $x$.

However, a student I am helping got this by saying $\ln x = 1$ gives $x = \mathrm{e}$ and at $x= \mathrm{e}$, the denominator $(1+ \ln x)^2 = 4$.

Hence $x= \mathrm{e}$.

Is this approach always correct or is it just luck here?

In general if I have $\frac{f(x)}{g(x)} = \frac{m(x)}{n(x)}$, can I solve it by finding the common solutions of $f(x) = m(x)$ and $g(x) = n(x)$?

[Edit: clearly not because if I have $\frac{x}{x+2} = \frac{1}{x+3}$, then $x= 1$ and $x+2 = x + 3$ don't give you anything..., so why does it work in this case?]

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  • $\begingroup$ you have just found one solution, namely $x=e$. But this method tells you nothing about whether or not there are more solutions. Also in your edit, "$x+2=x+3$ don't give you anything" is an incorrect statement. It tells you that THERE IS NO solution (I'm not sure if you meant this and were sloppy with the language or something else). So, what your example shows is that setting numerators equal and denominators equal is in general not a good strategy, because there might not be a common solution. $\endgroup$ – peek-a-boo Jun 26 at 22:31
  • $\begingroup$ Let $u = \frac{\log x}{1+\log x}$, your equation becomes $u(1-u) = \frac14$ which has a double root at $u = \frac12$. $x = e$ is the unique solution because the root is a "double" root. $\endgroup$ – achille hui Jun 26 at 22:37
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$\frac ab = \frac cd$ does not mean $a = c$ and $b = d$.

Here are two examples:

$\frac 13 = \frac 26$ but $1 \ne 2$ and $3\ne 6$. Likewise if $x =2$ we would have $\frac x{x+4} = \frac 13$. That does not mean $x=1$ and $x+4 =3$.

One (not advised) way of doing this is to notice that $\frac ab = \frac {\beta a}{\beta b}$ for all $\beta$ (excepts $0$) so there must be some $\alpha$ where we have $\frac ab = \frac {\alpha a}{\alpha c} = \frac cd$ and $a = \alpha c$ and $b=\alpha d$. We just don't have any reason to assume that $\alpha$ is equal to $1$.

So when $\frac 13 = \frac 26$ we have $1 = [\frac 12]*2$ and $3 = [\frac 12]*6$. And for $\frac x{x+4} = \frac 13$ we have $x = \alpha$ and $x+ 4 = 3\alpha$. (We can solve this as $\alpha = x$ we have $x +4 = 3x$ so $2x =4$ and $x=2$...)

But the better way is to do:

$\frac ab = \frac cd \implies ad = bc$. For our examples we get $\frac 13 = \frac 26 \implies 1*6 = 2*3$ which it does and $\frac x{x+4} =\frac 13 \implies 3x = x+4\implies x = 2$.

So in your problem you have:

$\frac{\ln x}{(1+ \ln x)^2} = \frac{1}{4}$ so that means

$4\frac {\ln x} = (1 + \ln x)^2$. It does not mean $\ln x = 1$ or that $(1+\ln x)^2 = 4$. It does mean that there is some constant $\alpha$ so taht $\ln x = \alpha$ and $1+\ln x)^2 = \alpha *4$ but .... that's just the same thing with a complicated and unnecessary variable thrown in.

To do this I'd just replace $\ln x$ with $w$ and have $4w = (1 + w)^2$ and solve for $w$.

$4w = 1 + 2w + w^2$

$0 = 1-2w + w^2$

$(w-1)^2 = 0$

$w-1 = 0$

$w = 1$

So $\ln x = 1$ so $x = e$.

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If $f(x) = m(x)$ and $g(x) = n(x)$ do have a common solution, then that solution is also a solution of $\frac{f(x)}{g(x)} = \frac{m(x)}{n(x)}$. The problem is that it is not guaranteed to be the only solution. An example where this is the case would be:

$$\frac{(x-1)}{2x-1} = \frac{(x-1)}{x(3x-2)}$$

The solution $x=1$ can be obtained using the method you proposed, but there is another solution, namely $x=1/3$.

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  • $\begingroup$ Where do the other solutions come from? I'm guessing cancellations etc? $\endgroup$ – PhysicsMathsLove Jun 26 at 22:34
  • $\begingroup$ @PhysicsMathsLove Look at the example I provided in my edit. $\endgroup$ – Haris Gušić Jun 26 at 22:39
  • $\begingroup$ @PhysicsMathsLove Consider $\dfrac{e^{-x^2}}{1-e^{-e}+e^{-x}}=e^{-e^2}$. $\endgroup$ – Gae. S. Jun 26 at 22:58
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Let $u = \ln x$.

Then,

$$ \frac{u}{(u^{2} + 1)^{2}} = \frac{1}{4}$$

from which it follows

$$ 4u = u^{2} + 2u + 1 $$

that is,

$$ u^{2} - 2u + 1 = (u - 1)^{2} = 0 $$

whence,

$$ u = 1 ; $$

Therefore, $x = e$ is the only solution.

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  • $\begingroup$ Let me know if I should delete my answer in time deference to Henry Lee. I've never seen anything quite like this before! $\endgroup$ – mlchristians Jun 26 at 22:53
  • $\begingroup$ To whom it may concern: Please up-vote Henry Lee's solution over mine. Thank you. $\endgroup$ – mlchristians Jun 26 at 23:30
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$$\frac{\ln x}{(1+\ln x)^2}=\frac{1}{4}$$ with $u=\ln(x)$ we get: $$\frac{u}{(1+u)^2}=\frac{1}{4}$$ $$4u=1+2u+u^2$$ $$u^2-2u+1=0\Rightarrow (u-1)^2=0$$ $$\therefore u=1$$ $$x=e^u\Rightarrow x=e,$$ This appear to be the only solution

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    $\begingroup$ Not that it matters, but $(u-1)^2 = 0$ implies only $u = 1$, not $u = -1$. $\endgroup$ – mechanodroid Jun 26 at 23:59
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    $\begingroup$ $u= -1$ isn't a solution. You have $(u-1)^2 = 0$ so $u-1 = \pm 0$ and $u = 1 \pm 0$. (In other words as $(u-1)^2$ is a perfect square it has a double root. $\endgroup$ – fleablood Jun 27 at 1:10
  • $\begingroup$ I've corrected it now, dont really know what I was thinking $\endgroup$ – Henry Lee Jun 27 at 8:27

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