12ln x = x^3/2
Let y=ln x, so 12y = x^3/2
Take logs of each side:
ln(12y) = 3/2ln x and substitute y=ln x:
ln(12y) = 3/2y
ln12 + ln y = 3/2y or 2.485 + ln y = 3/2y
Consider the function ln12 + ln y - 3/2y = 0 or consider the two functions f(y)=3/2y and g(y)=ln(12y) and consider what they look like as two functions on one graph. So we have a straight line of gradient 3/2 and a log curve. Where they intersect is the solution for y. If we put y=2 we get f(y)=3 and g(y)=ln24=3.18 approx., and when y=3, f(y)=9/2=4.5 and g(y)=ln36=3.58. In the first case f<g and in the second case f>g, so y must be between 2 and 3, because the straight line and curve cross somewhere between. We keep narrowing the search by trial and error and finally arrive at y=2.1744, approx., from which x can be calculated because if y=ln x, x=e^y, making x=8.797 approx.