How can you solve this equation?

Question

$$Q:{ 2 }^{ x }=4x$$

I tried to solve it and I got 4 by guessing . I wanna help me in finding algebraic solution .

Answer 1

You can algebraically rearrange this equation for hours and still get nowhere, because we cannot use algebra to solve it. Frustrating, isn't it?

In short, you are better off letting a computer solve this for you via the Lambert W-Function (Wolfram Alpha will do this). Alternatively, you can graph the two sides of the equation and find the approximate intersections. If you are dead set on doing it by hand, it will take quite a while, as the solution is transcendental (it will technically take an infinite amount of time to solve this problem numerically).

You were right to guess $x=4$. Another solution is $x\approx0.309907$. I obtained this from Wolfram Alpha.

Answer 2

Lambert solution
$$ 2^x=4x \\ e^{x\log 2} = 4x \\ 1=4xe^{-x\log 2} \\ \frac{-\log 2}{4}=(-x\log 2)e^{-x\log 2} \\ W\left(\frac{-\log 2}{4}\right) = -x\log 2 \\ \frac{-1}{\log 2}W\left(\frac{-\log 2}{4}\right) = x $$ Using the two real branches of $W$, we get two real solutions: $$ \frac{-1}{\log 2}W_0\left(\frac{-\log 2}{4}\right) = 0.3099069\dots \\ \frac{-1}{\log 2}W_{-1}\left(\frac{-\log 2}{4}\right) = 4.0000000\dots $$

Answer 3

Note that we cannot have a solution with $x<0$ because then LHS>0 but RHS<0 (where I using the standard abbreviations for left-hand side (here $2^x$) and right-hand side (here $4x$).

Notice that $2^0>4\cdot0$, but $2^1<4\cdot1$ and $2^5>4\cdot5$, so we expect one solution between 0 and 1 and another between 1 and 5, although so far we have ruled out any of (1) the possibility of more than one solution between 0 and 1, (2) the possibility of more than one solution between 1 and 5, (3) the possibility of solutions $\ge5$.

It is also easy to check that the only integer solution between 0 and 5 is $x=4$. So any solutions $>0$ and $<4$ are not integers.

This is tagged pre-calculus which makes further progress more difficult.

Notice that the rate of increase of the RHS is always the same. If we increase $x$ by $k$, then we increase RHS by $4k$. But the rate of increase of the LHS is continually increasing. If we increase $x$ by $k$ then we increase $2^x$ to $2^{x+k}=2^x2^k$ and hence by $$2^x(2^k-1)\ \ (*)$$ But $2^k>1$ for $k>0$. Moreover it is obvious that $2^x$ is strictly increasing, so $2^k-1$ is also, and hence for any fixed $k>0$ the increase $(*)$ is strictly increasing as $x$ increases.

This means that once $2^x$ overtakes $4x$ it always stays ahead, or in math language if $2^x_0>4x_0$ for some $x_0>0$, then $2^x>4x$ for all $x>x_0$. [Can you see why we need the caveat that $x_0>0$?] Clearly at $x=3$ we have $2^x<4x$. Moreover for $x>2$ the increase in the LHS if we increase $x$ by 1 is more than $2^2=4$ (from (*) above) whereas the increase in the RHS is 4. So it is clear that we have exactly one solution for $x>2$, which must be the one you found (at $x=4$).

So we have established everything except that there is just one solution in the range $0<x<2$. If you use numerical methods you can easily establish that the solution is $\approx0.30990693238069$. But at the moment I cannot see how to do more with the armoury limited to pre-calculus.