Find the coordinates of a point on the curve y=ln(e^x +e^-x)

the gradient given in 3/5

thanks in advance!

y=ln(e^x +e^-x)

differentiate ...

y' = [1/(e^x +e^-x)] [e^x - e^-x]

y' = [e^x - e^-x]/[e^x + e^-x]

gradient given in 3/5

3/5 = [e^x - e^-x]/[e^x + e^-x]

3[e^x + e^-x] = 5[e^x - e^-x]

3e^x + 3e^-x = 5e^x - 5e^-x

-2e^x = - 8e^-x

-2e^x = - 8/e^x

e^x = 4/e^x

(e^x)² = 4

e^x = ± 2

take positive sign ...
e^x = 2
x = ln(2)

take negative sign ...
e^x = -2
x = ln(-2) <---- not defined (i.e. log of negative number does not exist.

so for x=ln(2)
y=ln(e^ln(2) +e^-ln(2))
y=ln(e^ln(2) + 1/e^ln(2))
y=ln(2 + 1/2)
y=ln(5/2)

coordinates: (ln(2),ln(5/2))

Let the point be (x1, y1)
=> y1 = ln (e^x1 + e^-x1) ... (1)

dy/dx = (e^x - e^-x) / (e^x + e^-x)
=> 3/5 = (e^x1 - e^-x1) / (e^x1 + e^-x1)
=> 3e^x1 + 3e^-x1 = 5e^x1 - 5e^-x1
=> e^x1 = 4e^-x1
=> e^x1 = 2
=> x1 = ln2
Plubbing e^x1 = 2 in (1),
y1 = ln (2 + 1/2) = ln(5/2)

=> the required point is
[ln2, ln(5/2)]