How do I solve $3=\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}$ for $k$?

Question

How do I solve this for $k$?

$$3=\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}$$

I know $k$ can be estimated by graphing, but is there a way to solve using algebra? If so, how?

Answer 1

Let: $x = 5+\dfrac{k}{5+\dfrac{k}{5+\dfrac{k}{...}}} \implies x = 5 +\dfrac{k}{x} \implies x^2-5x=k$

Substitute this in the original equation: $3= \dfrac{k}{x} \implies 3 = \dfrac{x^2-5x}{x} \implies x = 8$

$x =8 \implies k = 24.$

Answer 2

Notice if it does equal anything (or if it even makes sense to talk of it) then

$3 = \color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}$

$8 = 5 +\color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}$

$\frac 1 8 = \frac {1}{5 +\color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}}$

$\frac k8 = \frac {k}{5 +\color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}}=3$

so

$\frac k8 = 3$

....

or $3 = \frac k{5 + \color{blue}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}}$

$3 = \frac k{5 + \color{blue}3}$

.....

The real question is does it equal anything and does it make sense to talk of it?

.....

Here's a case of something that we can't talk about equaling anything.

$2x = 2^{(2^{(2^{\dots)})})}$ Then $2x = 2^x$ so $x = 2$. Doesn't make sense and $2^{2^{2^{\dots}}}$ obviously gets bigger with each step and there's no way doing it "infinitely many times" would settle down to simply $4$.

Likewise $S = 1 + 2 + 4 + 8 + 16 + 32 + .....$

$2S = 2 + 4 + 8 +16+32 + 64 +..... $

$1 + 2S = 1 + 2 + 4 + 8 +16+32 + 64 +..... $

$1+2S = S$ so $S = -1$ ... uh, say what?

So we do have to ask does $3 = \color{green}{\frac{k}{5+\frac{k}{5+\frac{k}{5+\cdots}}}}$ make sense?

......

Which it will if the terms get close to and converge to somthing.

Let $k = 24$.

Then $\frac {24}{5 + 3} = 3$.

And $\frac {24}{5 + \frac {24}{5+3}} = \frac {24}{5+3} = 3$.

And $\frac {24}{5 + \frac {24}{5+\frac {24}{5+3}}}=\frac {24}{5+3}=3$

and so on... so in this case... it's okay.