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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?

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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 $x\approx 3.616686239$.

Edit

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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