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I have tried using the quadratic formula and I also attempted to solve it for $x$ but just ended up with $8x = 8x$. I can see on a graph that they do meet at a two points, but I can't figure out where they meet.

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Is this $\frac12(x^2-x)$ or $\frac1{2(x^2-x)}$? – Hagen von Eitzen Jan 6 '13 at 12:53
It is the first you mentioned, x^2/2 - x/2. I edited the title for clarity – Danny Birch Jan 6 '13 at 12:57
up vote 4 down vote accepted

One obvious solution is $x=1$, when both sides of $$\frac12(x^2-x)=x\log_b x$$ become $0$. Note that the right side of is defined only when $x>0$ so we shall assume so. Thus we are allowed to divide by $x$ and arrive at $$\frac12(x-1)=\log_b x$$ and after exponentiation $$\tag1x=\sqrt{b^{x-1}}.$$ While we don't find an explicit expression for the solution, $(1)$ can be used as a recursion formula for a sequence quickly converging to a solution. The value depends on which logarithm we use and can be obtained only as a numerical approximation this way as $x\approx 0.749228$ if $b=10$ and $x\approx 3.5128624$ if $b=e$.

EDIT: For some bases $b$, the inverse of $(1)$ is better suited to iterate towards the fixpoint, i.e. to let $x_{n+1}=1+2\log_bx_n$. For example, this way one obtains $x\approx 6.319722$ if $b=2$.

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Excellent answer, thanks! I did type lg for base 2 but I think it got corrected – Danny Birch Jan 6 '13 at 13:12
Sometimes its useful define $0\log0$ as $0$, because of its limit for $x\to0$ (for example in the Information Theory). That would give us solution $x=0$. – qmsource Jan 6 '13 at 13:36
What do you use as the initial number in the recursive formula? I can't seem to find a proper solution for base 2 – Danny Birch Jan 6 '13 at 17:26
@DannyBirch I added the solution for $b=2$, which needs a different iteration for convergence. By the way, I am used to $\lg=\log_{10}$, $\ln=\log_e$, $\operatorname{ld}=\log_2$ (and $\log$ is either any or $\ln$). – Hagen von Eitzen Jan 6 '13 at 17:55
@HagenvonEitzen Ah yes, it seems that is the ISO standard. Thanks for the edit :) – Danny Birch Jan 6 '13 at 18:01

Note that since $x\neq0$: $$x-1 = 2\log x$$ And so $x=1$ is a solution (by inspection). I'm afraid the second solution can't be obtained without using the Lambert W function, but numerically it's around $~3.5$.

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Great, I'll have a look at that thanks :D – Danny Birch Jan 6 '13 at 13:12

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