# how would one solve the following equation

How could the following equation be solved?

$$100x^2=2^x$$

This is as far as I have got: $$\ln(100x^2) = \ln(2^x)$$

• Your logarithm algebra is incorrect. You should have $\ln(100x^2) = \ln(100)+\ln(x^2) = \ln(100)+2\ln(x)$ and $\ln(2^x) = x\ln(2)$. This comes from the property of logs that $\log(a^b) = b\log(a)$ and $\log(xy) = \log(x)+\log(y)$ – graydad Feb 3 '15 at 20:04
• See this. – Aaron Maroja Feb 3 '15 at 20:14

The equation has multiple solutions, one of which is $$x = -2 \,\frac{ W(\ln(2)/20)}{\ln(2)} \approx 0.103658.$$
This problem can be approached in two different ways: using techniques for approximating solutions through Newton's Method, Bisection Method, Interpolation or with the omega function. I will consider the latter. $$100x^2=2^x \\\frac{x^2}{2^x}=\frac{1}{100}\\\sqrt{\frac{x^2}{2^x}}=\sqrt\frac{1}{100}\\\frac{x}{2^{\frac{x}{2}}}=\pm\frac{1}{10}\\x2^{-\frac{x}{2}}=\pm\frac{1}{10}\\W\left(\frac{-xln(2)}{2}e^{-\frac{xln(2)}{2}}\right)=W\left(\mp\frac{ln(2)}{20}\right)\\\frac{-xln(2)}{2}=W\left(\mp\frac{ln(2)}{20}\right)\\x_1=\frac{-2W\left(-\frac{ln(2)}{20}\right)}{ln(2)}\approx0.103658\\x_2=\frac{-2W\left(\frac{ln(2)}{20}\right)}{ln(2)}\approx-0.096704\\x_3=\frac{-2W_{-1}\left(-\frac{ln(2)}{20}\right)}{ln(2)}\approx14.3247$$