How to solve $n$ from $ 2(0.914)^{1-n}-(0.829)^{1-n}=1$? How to find the approximate value of $n$ from the following equation
$$ 2(0.914)^{1-n}-(0.829)^{1-n}=1$$
?
Clearly taking logarithmic function both side doesn't give us any easy equation.
I think some numerical approach would be appropriate here.
Any help is appreciated
 A: The function $f(x)=2\times 0.914\times 0.914^{-x} - 0.829 \times 0.829^{-x}$ satisfies $f(1)=2-1=1$. Moreover the function $g(x) = f(x)-1$ has its only maximum at $x\approx 0.6$, as we see by equating its derivative with zero, $$0=\frac{dg(x)}{dx}= - 2\times 0.914 \ \ln (0.914) \times 0.914^{-x} + 0.829 \ \ln (0.829) \times 0.829^{-x}$$ which yields $$\frac{2\times 0.914 \ \ln (0.914)}{0.829 \ \ln(0.892)} = \left(\frac{0.914}{0.829}\right)^x $$ which we abbreviate as $a=b^x$ and thus
$$x=\frac{\ln(a)}{\ln(b)}=:x_M \approx 0.57... .$$
We can show it is a maximum if we compute the second derivative.
The function $g$ thus has two zeros, i.e. two positions where $f(x)=1$, one is at $x_1=1$, the other at $x_2\approx 0.12 ... >0$.
This second solution is in contrast to the suggestion by RidB that the second solution might be integer and if it is, it is $0$.
A: First of all, let's find the graph of the function $f(n) = 2(0.914)^{1-n}- (0.829)^{1-n} = m$ with $n$ on $ x$-axis and m on $y$-axis.
This is the graph
And this is the zoomed in picture of the graph at the crucial point
As you can clearly see, the value of m is 1 only at $n=0,1$. It is close to 1 at $n=-1$ but not considerable.
Hence there are only 2 answers, if $n$ is an integer, $n = 0, 1.$
And, if n is a rational number then $n = [0, 1]$
Thanks.
