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Use the Pythagorean equation and simplify for ln(x)
(ln(3x))^2 = (ln(2x))^2+(lnx)^2)
(ln(3)+ln(x))^2 = (ln(2)+ln(x))^2 + ln(x)^2
(ln(x))^2+2ln(3)ln(x)+(ln(3))^2 = ln(x)^2+2ln(2)ln(x)+(ln(2))^2 + (ln(x))^2
Let y= ln(x), note that y> zero for a valid solution.
y^2+2ln(3)y + (ln(3))^2 = y^2+2ln(2)y+ (ln(2))^2 + y^2
y^2 -2y(ln(3)-ln(2))+(-ln(3)^2 + ln(2)^2) = 0
using the quadratic formula and lots of algebra
y = ln(3/2)+-sqrt(ln(3/2)ln9)) only the positive square root gives y>0
x = e^y which approximates to x = 3.854
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