Is it possible to solve for x, where ln(x), ln(2x) and ln(3x) form a legitimate right triangle?
From: Artofmathematics.com
To be the side of a triangle, ln(x) must be a positive number, so x>1
let y = ln(x)
a,b,c are y, ln(2)+y, ln(3)+y
For easier typing, let M = ln(2) and N = ln(3), note N>M
y^2 + (M+y)^2 = (N+y)^2
y^2 + M^2 + 2My + y^2 = N^2 + 2Ny + y^2
y^2 - 2(N - M)y - (N^2 - M^2) = 0
y = (N - M) ± sqrt[(N - M)^2 + (N^2 - M^2)]
y = (N - M) ± sqrt[(N^2 - 2MN + M^2 + N^2 - M^2]
y = (N - M) ± √2*√(N^2 - MN)
plug in M = ln(2) = 0.6931471805599453
N = ln(3) = 1.0986122886681098
y1 = (N - M) + √2*√(N^2 - MN)
= 1.349338987812199
y2 = (N - M) - √2*√(N^2 - MN)
= -0.5384087715958701 REJECT since y must be > 0
ln(x) = y1 = 1.349338987812199
x = e^y1 = e^1.349338987812199
x = 3.854876567948182
Side Lengths
ln(x) = ln(3.854876567948182)
ln(2x) = ln(7.7097531358516)
ln(3x) = ln(11.5646297037774)
side lengths squares of sides
1.34933898780639 1.82071570401438
2.04248616836634 4.17174974796781
2.4479512764745 5.99246545199315
1.82071570401438 + 4.17174974796781 = 5.99246545198219
which is accurate to eleven decimal places
Precise version of x:
x = e ^ ((ln(3)-ln(2)) + √2*√(ln(3)^2 - ln(2)ln(3)))
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Posted by Larry
on 2024-02-08 08:42:23 |
Solution
