Reverse an equation with ln and power

Question

I'm trying to solve for $x$ in the following equation: $\ln(y) = a \cdot (\ln(x)) ^ b + c$

$a = 0.0838 b = 2.6275 c = 0.2506$

but my results look bad. Can anybody show me his demonstration ?

Answer 1

Hint:

If $y=ax^b+c$ then $x=\left({\dfrac{y-c}{a}}\right)^{1/b}$ and you can then deal with your logs and antilogs.