Solve x in terms of y.
Question:
y = (e^x + e^-x)/(e^x - e^-x)
Answer:
x= 1/2 ln ((y+1)/(y-1))
I have the answer with me, but I don’t know how to solve the question step by step.
y = ( e^x + e^-x ) / ( e^x - e^-x )
e^-x = 1 / e^x
y = ( e^x + 1 / e^x ) / ( e^x - 1 / e^x )
Replace e^x with u
y = ( u + 1 / u) / ( u - 1 / u)
y = ( u ∙ u / u + 1 / u) / ( u ∙ u / u - 1 / u)
y = ( u^2 / u + 1 / u) / ( u^2 / u - 1 / u)
y = [ ( u^2 + 1 ) / u ] / [ ( u^2 - 1 ) / u ]
y = ( u^2 + 1 ) / ( u^2 - 1 )
Multiply both sides by ( u^2 - 1 )
y ∙ ( u^2 - 1 ) = u^2 + 1
y ∙ u^2 - y = u^2 + 1
Subtract u^2 to both sides
y ∙ u^2 - y - u^2 = u^2 + 1 - u^2
y ∙ u^2 - y - u^2 = 1
Add y to both sides
y ∙ u^2 - y - u^2 + y = 1 + y
y ∙ u^2 - u^2 = y + 1
u^2 ( y - 1 ) = y + 1
Divide both sidess by y - 1
u^2 = ( y + 1 ) / ( y - 1 )
Replace u = e^x in this equation
( e^x )^2 = ( y + 1 ) / ( y - 1 )
Take the natural logarithm of both sides
2 ∙ ln ( e ^ x) = ln [ ( y + 1 ) / ( y - 1 ) ]
since ln ( e ^ x) = x
2 ∙ x = ln [ ( y + 1 ) / ( y - 1 ) ]
Divide both sides by 2
x = ( 1 / 2 ) ln [ ( y + 1 ) / ( y - 1 ) ]
y = (e^x + e^-x)/(e^x - e^-x) = (e^(2x)+1)/(e^(2x)-1)
e^(2x)+1 = e^(2x)*y - y
e^(2x)(y-1) = y+1
e^(2x) = (y+1)/(y-1)
2x = ln (y+1)/(y-1)
x = 1/2 ln (y+1)/(y-1)
To solve for x in terms of y, we need to isolate x on one side of the equation.
1. Multiply both sides of the equation by e^x - e^-x to get rid of the denominator:
y * (e^x - e^-x) = e^x + e^-x
2. Expand both sides of the equation:
y * e^x - y * e^-x = e^x + e^-x
3. Move all terms containing e^-x to the left side of the equation, and all terms containing e^x to the right side:
y * e^-x - e^x = e^x - y * e^x + e^-x
4. Combine similar terms:
y * e^-x - e^x = (1 - y) * e^x + e^-x
5. Move all terms containing e^x to the left side of the equation, and all terms containing e^-x to the right side:
y * e^-x - (1 - y) * e^x = e^-x + y * e^x
6. Factor out e^x from the left side:
e^x * (y * e^-2x - (1 - y)) = e^-x + y * e^x
7. Divide both sides of the equation by (y * e^-2x - (1 - y)):
e^x = (e^-x + y * e^x) / (y * e^-2x - (1 - y))
8. Multiply both sides of the equation by e^2x:
e^x * e^2x = (e^-x + y * e^x) / (y * e^-2x - (1 - y)) * e^2x
9. Simplify:
e^3x = (e^-x + y * e^x) / (y * e^-2x - (1 - y))
10. Take the natural logarithm of both sides:
ln(e^3x) = ln((e^-x + y * e^x) / (y * e^-2x - (1 - y)))
11. Simplify:
3x = ln((e^-x + y * e^x) / (y * e^-2x - (1 - y)))
12. Divide both sides of the equation by 3:
x = (1/3) * ln((e^-x + y * e^x) / (y * e^-2x - (1 - y)))
Now, we have the solution for x in terms of y:
x = (1/3) * ln((e^-x + y * e^x) / (y * e^-2x - (1 - y)))
However, it seems that the provided answer is slightly different:
x = (1/2) * ln((y + 1) / (y - 1))
Please double-check the answer to make sure it is accurate.