I am trying this math to solve.the equation is like this:
$2^x=4x$
Here $x=4$ should be the answer.
But How can I derive it in a mathematical way?
I am failed to solve it my general algebra.How can I use logarithm here to solve it?
Thanks in Advance.
We have:
$$2^x=4x$$ $$\frac14=x\cdot0.5^x$$ $$\frac14=xe^{x\ln0.5}$$ $$\frac1{4\ln0.5}=x\ln0.5e^{x\ln0.5}$$
Taking $u=x\ln0.5$:
$$ue^u=\frac1{4\ln0.5}$$
This equation cannot be solved in elementary terms. However, the Lambert W function $\operatorname W(z)$ is defined such that it is the inverse of $ze^z$. Thus:
$$u=\operatorname W\left(\frac1{4\ln0.5}\right)$$ $$x\ln0.5=\operatorname W\left(\frac1{4\ln0.5}\right)$$ $$x=\frac{\operatorname W\left(\frac1{4\ln0.5}\right)}{\ln0.5}$$
This gives us $x\approx0.309907$.
Alternatively, if you want to use repetitive methods to calculate $2^x=4x$, then we define $f(x)=2^x-4x$. Now, we need to find $f(x)=0$.
The derivative of $f(x)$ is:
$$\frac d{dx}\left[2^x-4x\right]$$ $$=\frac d{dx}\left[2^x\right]-\frac d{dx}[4x]$$ $$=\ln2\cdot2^x-4$$
By Newton's method, we have
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{2^{x_n}-4x}{\ln2\cdot2^{x_n}-4}$$
Now given any initial guess $x_0$ we can use this iterative function to gain an approximation.