Solve cosh^-1(2x)=sinh^-1x?

cosh^-1(2x) = sinh^-1(x)

recall that cosh^-1(x) = ln [ x + √(x^2 - 1) ] and sinh^-1(x) = ln [ x + √(x^2+1) ]

so ln [2x + √(4x^2 - 1) ] = ln [ x + √(x^2+1) ]

2x + √(4x^2 - 1) = x + √(x^2 + 1)

√(x^2 + 1) - √(4x^2 - 1) = x

squaring

x^2 + 1 + 4x^2 - 1 - 2√(x^2 + 1)√(4x^2 - 1) = x^2

2√(x^2 + 1)√(4x^2 - 1) = 4x^2

=> √(x^2 + 1)√(4x^2 - 1) = 2x^2

again squaring

=> (x^2 + 1)(4x^2 - 1) = 4x^4

4x^4 + 3x^2 - 1 = 4x^4

3x^2 - 1 = 0

3x^2 = 1

x^2 = 1/3

x = 1/√3 (ignoring invalid negative value)

x = (1/3)√3

is this cosh inverse or reciprocal of cosh ?

if it was reciprocal you could have simply said cosh(2x) = sinh(x) so it was a silly doubt i agree

anyway cosh^-1(2x) is inverse of cosh(2x) = ln[2x+ sqrt (4x^2-1) ]

inverse of sinh x = ln[x+ sqrt (x^2+1) ]

we need to solve ln[2x+ sqrt (4x^2-1) ] = ln[x+ sqrt (x^2+1) ]

2x+ sqrt (4x^2-1) = x+ sqrt (x^2+1)

x+ sqrt (4x^2-1) = sqrt (x^2+1)

x^2 + 4x^2 - 1 + 2xsqrt(4x^2-1) = x^2 + 1

4x^2-2 = -2xsqrt(4x^2-1)

16X^4 -16x^2 +4 = 4x^2 (4x^2-1)

16x^4 -16x^2 +4 = 16x^4 -4x^2

12x^2 = 4

x= 1/root3

i hope i havent missed something and made a longer approach to a simpler problem

let me know

all the best

y = coshֿ¹(2x)

2x = cosh(y)

x = cosh(y)/2

y = sinhֿ¹(x)

x = sinh(y)

cosh(y)/2 = sinh(y)

2cosh(y) = 4sinh(y)

e^(y) + e^(-y) = 2e^(y) - 2e^(-y)

3e^(-y) = e^(y)

e^(2y) = 3

e^(y) = √(3)

y = (1/2)ln(3)

x = sinh(y)

= sinh[(1/2)ln(3)]

= (1/2)e^[(1/2)ln(3)] - (1/2)e^[-(1/2)ln(3)]

= (1/2)√(3) - (1/2)/√(3)

= √(3)/3