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How can I solve the equation $\dfrac x9 = \log_2x$ ?

Amzoti
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dev_nut
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  • Do you have thoughts on the problem and can share those? What have you tried? Regards – Amzoti Sep 02 '13 at 02:35
  • I have tried taking out the logarithms but that just makes it into an equation with 2^(x/9) = x. I'm not sure how to solve that either. So, I thought I'm missing some fundamental technique. – dev_nut Sep 02 '13 at 02:40
  • Hint: http://en.wikipedia.org/wiki/Lambert_W_function (example 4) – Amzoti Sep 02 '13 at 02:43

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If you plot the graphs of $y=\frac{x}{9}$ and $y=\log_2 x$, you will see that there are two solutions: $1 < x_1 < 2$ and $5 < x < 6$. You can improve the accuracy by using, e.g., Newton's method.

Doctor Dan
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I was having trouble with a similar equation, $x^2+ln(x) = y$, and I found the Lambert W function, that is useful to solve equations in which the unknown appears both outside and inside an exponential or a logarithm.
Basically $W=f(x)^{-1}$, where $f(x) = xe^x$, so $W(xe^x) = x$.

We have $\frac{x}9=log_2(x)$ that can be written as $\frac{x}9=\frac{ln(x)}{ln(2)}$ so we have e as the base, and then we can rewrite it as $\frac{ln(2)}{9}=\frac{ln(x)}{x}$, then we solve. $\require{enclose}$ $$\frac{ln(2)}{9}=\frac{ln(x)}{x}$$ $$\frac{ln(2)}{9}=\frac{ln(x)}{e^{ln(x)}}$$ $$\frac{ln(2)}{9}={ln(x)}{e^{-ln(x)}}$$ $$-\frac{ln(2)}{9}=-{ln(x)}{e^{-ln(x)}}$$ $$W(-\frac{ln(2)}{9})=W(-{ln(x)}{e^{-ln(x)}})$$ $$W(-\frac{ln(2)}{9})=-ln(x)$$ $$\bbox[5px,border:2px solid black]{ x_1=e^{-W(-\frac{ln(2)}{9})} ,\; x_2=e^{-W_{-1}(-\frac{ln(2)}{9})} }$$

You can check the result in WolframAlpha to see the exact values. Hope this was useful to someone looking for ways to solve this type of equations.

Santiago Chaves
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