The curves are $$ y = \log_n |x|$$ $$ y = mx^2 + k$$
For $0<n<1$ and $m<0$, what are the conditions of k for solutions to $\log_n |x| = mx^2 + k$?
I tried to apply the logic that for solutions, the integral area between the curves must be > 0, from where after simplification I got $$ (b-a)[(b^2 + ab + a^2)m*ln(n) + 3k*ln(n) + 3] > ln\Biggl(\frac{|b|^{3b}}{|a|^{3a}}\Biggl)$$ where a, -a and b, - b are the points of intersection and |b|>|a|
Now, there can be three cases:
a and b both over x-axis $\implies$ k > 0 $\implies$ R.H.S. > 1
b over x axis and a below it $\implies$ k>0 $\implies$ R.H.S. > 1
a and b both below x axis $\implies$ 1>R.H.S.>0$\implies$ |ab|>1
Therefore, L.H.S. in all the three cases > 0 (wherein we can exclude the (b-a) part as it is always positive)
I'm stuck at this point. Help? $$----$$ For solving further, I had replaced ln(n) with $\frac{(ln|ab|)}{m(a^2 + b^2) + 2k}$