Hyperbolic Function Example Question (notes)

Solve artanh(x^2-1/x^2+1) = ln2

• Method 1: We have to remember this is artan, not artanx. So we sub (x^2-1/x^2+1) into x values for artanx=ln((x+1/x-1)/2):
• 0.5 x ln(((1+(x^2-1/x^2+1))/(1-(x^2-1/x^2+1))) = ln2 , Now multiply by x^2+1 to remove double fraction: 0.5 x ln(((x^2+1)+(x^2-1))/(x^2+1)+(x^2-1))) = ln2. This cancels to give 0.5 x ln(x^2)=ln2. Now we take logs: ln(x^2)^(1/2) = ln2 , ln|x|=ln2 , e^ln|x| = e^ln2 , |x|=2 , so x=+/- 2
• ORRR
• Method 2: Multiply both sides by tanh to and use exponential terms to work out:
• (x^2-1)/(x^2+1) = tanh(ln2) , (x^2-1)/(x^2+1) = (e^ln2 - e^-ln2)/(e^ln2 + e^-ln2). We can rewrite e^-ln2 as e^ln2-1 which is the same as e^ln(1/2) which is the same as 1/2. So e^ln2 = 2 , so (x^2-1)/(x^2+1) = (2-(1/2)) / (2+(1/2)) , (2-(1/2)) / (2+(1/2) is the same as 3/5 , so (x^2-1)/(x^2+1)=3/5. Now we solve this by cross multiplying to get 2x^2 = 8 , so x=+/- 2 .