I have

$ 4^x - 4^\sqrt x+1 = 3*2^{x+\sqrt x}$

i have tried simplifying that equation, but that left me with a $ \sqrt x$ and an $ x$ in the exponential, which are two different values... and I am not even sure i got it right, but anyway:

$ 2^{x-2-\sqrt x} - 2^{-x} = \frac 34$

can you please help me out?


This looks to be an highly transcendental function and, more than likely, numerical methods would be required.

So, consider that you are looking for the zero of function $$f(x)=4^x - 4^\sqrt x+1 - 3*2^{x+\sqrt x}$$ We know $x\geq 0$. Let try inspection $$\left( \begin{array}{cc} x & f(x) \\ 0 & -2. \\ 1 & -11. \\ 2 & -22.0847 \\ 3 & -25.7636 \\ 4 & 49. \end{array} \right)$$ So, there is a root between $3$ and $4$.

The simplest would be to use Newton method which, starting from a "reasonable" guess $x_0$, will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

Let us apply the method $$\left( \begin{array}{cc} n & x_n \\ 0 & 3.50000 \\ 1 & 3.63419 \\ 2 & 3.61702 \\ 3 & 3.61669 \end{array} \right)$$ which is the solution for six significant figures.

Being patient and making more iterations (this would not be of any interest at all), you could arrive at $x\approx 3.616686239$.


After your simplication, which is good, we are let with the problem of the zero of

$$g(x)=2^{x-2-\sqrt x} - 2^{-x} - \frac 34$$ the plot of which being slightly nicer than the plot of $f(x)$ but the same problem and then same methods.


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