0
$\begingroup$

If given $$14^x + 23^{2x} = k$$

For all $k \in \mathbb{N}$. How to arrive at $x$?

Thanks in advance.

$\endgroup$
  • $\begingroup$ I doubt there is a clear algebraic way to handle this for general $k$. The function $f(x)=14^x+23^{2x}$ increases montonically so there is a unique solution for any natural number $k$ ...numerical methods will find it, but I doubt there will be a reasonable closed formula. Of course, I might have that wrong...but checking the form of the solution for modest $k$ doesn't suggest much. $\endgroup$ – lulu Jan 6 '17 at 14:33
  • 1
    $\begingroup$ Alas, logs don't work well with sums. That is to say, $\log (a+b)$ isn't any sort of nice function of $\log a,\log b$. $\endgroup$ – lulu Jan 6 '17 at 14:38
  • $\begingroup$ The answer to this question will be a set of values of $x$, and not just one value, right? $\endgroup$ – Shreyas S Jan 6 '17 at 14:38
3
$\begingroup$

I suggest using Newton-Raphson method. It's a numerical method, but very efficient.

The standard formulation is this:

To solve $f(x)=0$, start with an estimate $x_0$

Calculate $x_1=x_o-\dfrac {f(x_0)}{f'(x_0)}$

Continue to calculate $x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)}$

You should get a sequence of values $x_0, x_1, ... , x_n$ that converge to a root.

Note that this can not tell you how many roots there are - if you suspect that there more than one root, then you will have to take a number of different starting points.

In your case the equation $14^x+23^{2x}=k$ can be rewritten as $14^x+23^{2x}-k=0$, so $f(x)=14^x+23^{2x}-k$

This can be rewritten as $f(x)=e^{(\log 14)x}+e^{(2\log 23)x}-k$

Hence $f'(x)=(\log 14)e^{(\log 14)x}+(2\log 23)e^{(2\log 23)x}$

So $x_{n+1}=x_n-\dfrac {e^{(\log 14)x_n}+e^{(2\log 23)x_n}-k}{(\log 14)e^{(\log 14)x_n}+(2\log 23)e^{(2\log 23)x_n}}$

$\endgroup$
2
$\begingroup$

Fixed point iteration:

$$23^{2x}=k-14^x$$

$$2x=\log_{23}(k-14^x)$$

$$x=\frac12\log_{23}(k-14^x)$$

So for some $x_0\approx x$, we can use the following:

$$x_{n+1}=\frac12\log_{23}(k-14^{x_n})$$

For example, with $k=13$ and $x_0=0.3$,

$x_1=\frac12\log_{23}(13-14^{0.3})=0.3793470222$

$x_2=0.37156366654$

$x_3=0.372419725772$

etc. And $x=\lim_{n\to\infty}x_n$.

$\endgroup$
  • $\begingroup$ Yes, but very slow convergence for small $k$ $\endgroup$ – tomi Jan 6 '17 at 17:46
  • $\begingroup$ @tomi I agree, but the other answer took Newtons method, so... $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 17:47
0
$\begingroup$

As already said in answers and comments, Newton method is probably the best and simplest way to find the zero of $$f(x)=14^x + 23^{2x} - k$$ From a practical point of view, I would suggest to use instead $$g(x)=\log \left(14^x+23^{2 x}\right)-\log(k)$$ which is much better conditioned (just by curiosity, plot $14^x + 23^{2x}$ and $\log \left(14^x+23^{2 x}\right)$).

As you said in a comment, using $f(x)$, the convergence is quite slow for small values of $k$.

For illustration purposes, let us use both functions for $k=0.001$ starting using $x_0=0$.

For $f(x)$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & -0.224353505 \\ 2 & -0.490433809 \\ 3 & -0.805641173 \\ 4 & -1.156949545 \\ 5 & -1.520584615 \\ 6 & -1.876683719 \\ 7 & -2.201538319 \\ 8 & -2.453972852 \\ 9 & -2.586800792 \\ 10 & -2.616322943 \\ 11 & -2.617534838 \\ 12 & -2.617536781 \\ \end{array} \right)$$

while, for $g(x)$, they would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & -1.706142182 \\ 2 & -2.615735506 \\ 3 & -2.617536780 \\ 4 & -2.617536781 \\ \end{array} \right)$$

The same would apply to large values of $k$.

For $k >0 \in \mathbb{N}$, the graph of $\log \left(14^x+23^{2 x}\right)$ is almost a straight line.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.