# How to smoothen a contour plot?

I want to smoothen the following contour.

ContourPlot[
10^b - (10^180/(10^m  10^23)^6)^(1/4)/Sqrt[ (((10^36)^(3/2))/(
10^m  10^23 Sqrt[ 100]))] == 0, {m, -3, 15}, {b, -25, -2},
ContourStyle -> {{Purple, Thickness[.01]}}]


Is there a way to do so?

Try the option MaxRecursion

ContourPlot[
10^b - (10^180/(10^m 10^23)^6)^(1/4)/Sqrt[(((10^36)^(3/2))/(10^m 10^23Sqrt[100]))] == 0,
{m, -3, 15}, {b, -25, -2},
ContourStyle -> {{Purple, Thickness[.01]}},
MaxRecursion -> 5]


10^b - (10^180/(10^m 10^23)^6)^(1/4)/Sqrt[(10^36)^(3/2)/(
10^m 10^23 Sqrt[100])] == 0 // FullSimplify


get

10^b == (10^(-6 (-7 + m)))^(1/4)/Sqrt[10^(30 - m)]


Solve m

m /. Solve[10^b == (10^(-6 (-7 + m)))^(1/4)/Sqrt[10^(30 - m)], {m}]


get

{(-9 Log[10] - 2 Log[10^b])/(2 Log[10])}


Plot it

Plot[(-9 Log[10] - 2 Log[10^b])/(2 Log[10]), {b, -25, -2}]


This is m=f[b] , change to b=f[m]

Plot[InverseFunction[(-9 Log[10] - 2 Log[10^#])/(2 Log[10]) &][
m], {m, -3, 15}]


And after simplify, we get

Plot[1/2 (-9 - 2 m), {m, -3, 15}]


😂，such a simple expression.

• Simplify[10^b - (10^180/(10^m 10^23)^6)^(1/4)/ Sqrt[(((10^36)^(3/2))/(10^m 10^23 Sqrt[100]))], Assumptions -> -25 <= b <= -2 && -3 <= m <= 15] Commented Apr 12, 2021 at 8:40