$$1.02^n = n^{100}$$

Its a question of where does this exponential function meet the polynomial function?


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    $\begingroup$ Welcome to Math StackExchange. Learning the culture here can sometimes be a bit brutal. If a questioner does not include, along with the question, a description of what he/she has already tried, the question might be down-voted. Perhaps you could add a few words describing what you have already tried? $\endgroup$ – John Wayland Bales May 14 '17 at 18:28

First, this equation does not involve polynomials, since $n$ certainly will not be a positive integer. Accordingly, I will use $x$ in place of $n$.

First, one should take the logarithm of both sides:


Next define a function

$$ f(x)=100\ln(x)-x\ln(1.02) $$

Then we need to find the value of $x$ satisfying the equation

$$ f(x)=100\ln(x)-x\ln(1.02)=0$$

We can guess that the value of $x$ will be close to $1$ so we use Newton's method with an initial guess $x_0=1$.

In Newton's method one makes an initial guess for the value of $x$ and uses that guess to produce a better guess according to the formula

$$ x_{k+1}= x_k-\frac{f(x_k)}{f^\prime(x_k)}\tag{1}$$

With a first guess of $x_0=1$ we find that $f(1)=-\ln(1.02)$, and since

$$f^\prime(x)=\frac{100}{x}-\ln(1.02)\tag{2} $$

we find that


Substituting these values into equation $(1)$ above give a second guess of

$$ x_1=1-\frac{-\ln(1.02)}{100-\ln(1.02)}\approx1.00198 $$

If this is not sufficient precision, you may find $x_2$.


Here is a desmos graph of where the function crosses the $x$-axis: graph of function

Now let us look again at equation $(2)$ for $f^\prime(x)$. We see that $f$ will have slope $0$ when $x=\dfrac{100}{\ln(1.02)}\approx5050$. So for $x>5050$ the graph of $y=f(x)$ is decreasing indicating the possibility of another solution greater than $5050$. In fact the second derivative

$$ f^{\prime\prime}(x)=-\frac{100}{x^2} $$

tells is the graph is concave down guaranteeing that there is another solution larger than $5050$.

Equation $(2)$ tells us that the graph is very steep at $x\approx1$ but very flat (horizontal) at $x\approx5050$ so Newton's method converged very fast at $x_0=1$ but will converge much slower if we pick $x_0=5050$.

For this I resorted to a graphing tool, desmos.com and set a wide range for the $x$-axis: the interval $[5000,100000]$.

[1]: https://i.stack.imgur.com/wtnkF.png

This tells us that $x_0=55000$ is a much better first guess than $5050$.

Substituting into equation $(1)$ gives

$$ x_1=55131.5$$

Finally, by taking natural logarithm of both sides of the original equation

$$ (1.02)^x=x^{100} $$

I automatically excluded any negative solutions. To include those let

$$ g(x)=x^{100}-1.02^x \tag{3}$$

$$ g^\prime(x)=100x^{99}-\ln(1.02)1.02^x \tag{4}$$

A desmos.com graph of $g$ looks like this:

graph of g

indicating we should look for a solution to equation $(3)$ beginning with $x_0=-1$. This will lead to the negative solution $−0.999802$.

  • $\begingroup$ Thanks for the answer, but i checked, n is 55131.5 as well as an x3 solution. How that comes out? $\endgroup$ – Ádám Felméry May 14 '17 at 19:18
  • $\begingroup$ I forgot to show you how to check for other solutions. I will add that to my answer. $\endgroup$ – John Wayland Bales May 14 '17 at 19:40

HINT: by the Newton Raphson method can be found $$1.0002,-0.999802,55131.5$$

  • $\begingroup$ Isn't it possible to solve by simple ways, or using Lambert? $\endgroup$ – Ádám Felméry May 14 '17 at 18:44

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