Product of logartithms equation.

Question

Please help me whilst I do a few simple school tasks. I found this one, which is unbreakable for me. I will appreciate any help.

$$\log_2{x}=\frac{4}{\log_2 x-3}$$

I moved only with the fact that $4= \log_2 16$. I have no idea what to do next.

Answer 1

As far as I know, this doesn't have an elementary solution. The best I can get using numerical methods is $x \approx 5.94051$. Was this an exercise in numerical methods?

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If you meant $\log_2 x - 3$, then it can be solved the following way:

\begin{align} \log_2 x \cdot (\log_2 x - 3) &= 4 \\ (\log_2 x)^2 - 3 \log_2 x - 4 &= 0 \end{align}

Using a substitution $u = \log_2 x$ to get the above as $$u^2 - 3u - 4 = 0$$ This has solutions $u = -1$ and $u = 4$. The first yields $$\log_2 x = -1 \iff x = \frac{1}{2}$$ and the second yields $$\log_2 x = 4 \iff x = 16$$

Answer 2

Did you mean $\log_2(x) - 3$ ?

You get x = 1/2 and x = 16

http://www.wolframalpha.com/input/?i=ln+x+%2F+ln+2+%3D+4+%2F+%28ln%28x%29%2Fln2+-+3%29