Can one solve for $x$ in the equation $8^x=16x$

Question

The question is whether or not it is possible to solve for $x$ in the equation $8^x=16x$.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/ypio1.jpg any assistance will be appreciated

Answer 1

$$8^x=16x$$ $$e^{x\ln(8)}=16x$$ $$e^{-x\ln(8)}=\frac{1}{16x}$$ $$xe^{-x\ln(8)}=\frac{1}{16}$$ $$-x\ln(8)e^{-x\ln(8)}= -\frac{1}{16}\ln(8)$$ Let $\quad z=-x\ln(8)\quad$ and $\quad C=-\frac{1}{16}\ln(8)$ $$ze^z=C$$ This is the usual Lambert's equation which solutions are : $$z=W(C)$$ $W$ is the Lambert W function. http://mathworld.wolfram.com/LambertW-Function.html

$$x=-\frac{z}{\ln(8)}= -\frac{1}{\ln(8)}W\left(C\right)$$ $$x=-\frac{1}{\ln(8)}W\left(-\frac{1}{16}\ln(8)\right)$$

The Lambert W function is multi-valuated. The real solutions are commonly noted $W_0(X)$ and $W_{-1}(X)$. Finally the two real solutions are : $$x=-\frac{1}{\ln(8)}W_0\left(-\frac{1}{16}\ln(8)\right)\simeq 0.0727$$ $$x=-\frac{1}{\ln(8)}W_{-1}\left(-\frac{1}{16}\ln(8)\right)\simeq 1.54142$$ http://m.wolframalpha.com/input/?i=-lambertw%280%2C-ln%288%29%2F16%29%2Fln%288%29

http://m.wolframalpha.com/input/?i=-lambertw%28-1%2C-ln%288%29%2F16%29%2Fln%288%29

Of course, instead of using this special function, in practice the numerical approximate solutions are directly computed thanks to numerical method of calculus (Newton-Raphson for example).

Answer 2

Well, the most general problem is (when $\text{c}\ne0\space\wedge\space\text{c}\ne1$):

$$\text{a}+\text{b}\cdot\text{c}^x=\text{d}\cdot x\space\Longleftrightarrow\space x=\frac{\text{a}}{\text{d}}-\frac{\text{W}_\text{n}\left(-\frac{\text{b}}{\text{d}}\cdot\text{c}^\frac{\text{a}}{\text{d}}\cdot\ln\left(\text{c}\right)\right)}{\ln\left(\text{c}\right)}\tag1$$

Where $\text{W}_\text{k}\left(\text{z}\right)$ is the analytic continuation of the Product Log Function and $\text{n}\in\mathbb{Z}$.

---

So, in your problem:

$$0+1\cdot8^x=16\cdot x\space\Longleftrightarrow\space x=\frac{0}{16}-\frac{\text{W}_\text{n}\left(-\frac{1}{16}\cdot8^\frac{0}{16}\cdot\ln\left(8\right)\right)}{\ln\left(8\right)}=-\frac{\text{W}_\text{n}\left(-\frac{\ln\left(8\right)}{16}\right)}{\ln\left(8\right)}\tag2$$

Answer 3

The solution given using Lambert’s function is very elegant. However, if you are interested in obtaining the solution using numerical methods, you can follow this:

Let $f(x) = 8^x-16x$. This a smooth function.

It's obvious from the graph of $f(x)$ that it cuts x-axis, and hence you are guaranteed a solution. In fact, you can see it has two solutions.

Otherwise, use IVT to verify this: $f(1) = -6$ and $f(0) = 1$ and $f(1) = -6$ and $f(2) > 0$

By IVT, you are guaranteed a point $x_{sol} \in (0,1)$ and one more in $(1,2)$ such that $f(x_{sol})=0$. Since the function is monotonic ouside $(0,1)$ you will not get anymore solution points. )

You can use Newton's method to find the solution $x_{sol}$ \in (0,1) as follows:

Choose an initial point $x_0$ for Newton's method in $(0,1)$.

$df(x) = 8^x \ln 8-16$.

$|df(a)-df(b)| \leq \ln 8 |8^a-8^b|$ $\forall$ $a,b \in (0,1)$,

$\implies |df(a)-df(b)| \leq \ln 8 |a-b|$

Thus, $df(x)$ is Lipschitz in $(0,1)$ with a Lipschitz ratio $M = \ln 8$ in $(0,1)$,

Let $x_{0} \in (0,1)$

You can show that, $|f(x_0)M df^{-1}(x_0)^2| \leq 0.5$

and hence you are guaranteed to converge to a solution $x_{sol} \in (0,1)$ by Katorovich's theorem.

If you want the other solution point in $(1,2)$, choose $x_{0} \in (1,2)$, and Kantorovich’s theorem guarantees convergence of Newton’s method to $x_sol \in (1,2)$