HOW WILL THE LATEX BRAGGER DIE AGAIN?

We can first notice that the expression $4^x-2^x$ contains powers of two, in which we can simplify $4$ into $2^2$

$$4^x - 2^x = 56$$
$$2^{2^x} - 2^x = 56$$
$$2^{2\cdot x} - 2^x = 56$$
$$2^{2x}-2^x=56$$

We can then use our past property about how $x^y\cdot x^z=x^{y+z}$ to separate $2^{2x}$ into $2^2\cdot 2^x$

$$2^{2x}-2^x=56$$
$$2^2\cdot 2^x-2^x=56$$

Letting $2^x$ be $y$ will make things easier to see.
$$2^2\cdot y-y=56$$
$$4y-y=56$$
$$3y=56$$
$$y=\frac{56}{3}$$

Plugging $y$ back in gives us:

$$2^x=\frac{56}{3}$$

We can separately visualize a graph for both sides of the equal to see how many solutions there are for $x$.

$2^x$ would look something like the green line while 56/3 would look something like the blue line. We see that the two graphs meet and will only meet at one point as green will keep on increasing and won't ever come back down. The other side from line green will also only get closer to reaching 0 meaning there is only one point of intersection and $\boxed{\boxed{\text{one solution}}}$ to the equation $4^x-2^x=56.$


https://latex.artofproblemsolving.com/miscpdf/odytuuhf.pdf?t=1566149907681






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