How to solve $2^x+e^x=400$ This should be pretty easy, I know. It involves logs, but then there's this 400. So logs of what? And since $2^x$ and $e^x$ are different things, I can't substitute the values and solve as a second degree equation. Thank you all in advance.
 A: $$e^x(1+(\frac{2}{e})^x)=400$$
take the log
$$x+\log(1+(\frac{2}{e})^x)=\log 400$$
$$x=\log \frac{400}{1+(\frac{2}{e})^x}$$
then use Fixed-point Iteration Method by selecting $x=1$ to get new value of $x=5.4400198...$
repeat this many times to get 
$$x=5.837229692...$$ 
A: I don't see how you will get an algebraic solution.  To get a numeric one, you can just ask Alpha and find $x \approx 5.83723$  To find that yourself with a simple calculator, you can think that $2^x$ is probably small compared to $e^x$,  write the equation as $e^x=400-2^x, x=\log(400-2^x)$ and make it a fixed-point iteration:  $x_{i+1}=\log (400-2^{x_i})$, start with $x_0=0$ and iterate to convergence. I get convergence in Excel in 7 steps.
A: an algebraic solution doesn't exist, by a numerical method we get $$x\approx 5.8372296916570249573$$
A: You can rewrite the equation as follows. We have:
$$2^x = \exp\left[\log(2^x)\right] = \exp\left[x\log(2)\right]= \exp(x)^{\log(2)}$$
The equation can thus be rewritten as:
$$y^{\log(2)} + y = 400$$
where $y = \exp(x)$. This is not an algebraic equation, but in some sense it's close to one. If you approximate $\log(2)$ by a rational number $\frac{p}{q}$ and substitute $y = z^q$ then you get the algebraic equation
$$z^p + z^q = 400$$
