i John,

Multiply both sides by z/y to obtain

xz/y = ln[(1+a²)/(1-a²)]

Apply the exponential function t both sides so that the equation becomes

e^xz/y = e^{ln[(1+a2)/(1-a2)]} = (1+a²)/(1-a²)

Now solve for a.

I hope this helps,

Penny

John wrote back

Hi penny,

thanks for the answer with natural log.

I am still stuck with this question as I was stuck at the same point that you left off.

i.e. how to make 'a' subject of this equation?

e^(xz/y) = (1+a^2)/(1-a^2)

Thanks in advance!

John,

Let me make the notation a little easier by writing k = e^xz/y then

k = (1+a²)/(1-a²)

Multiply each side by (1 - a²) and the equation becomes

(1 - a²) k = (1 + a²)

Simplify, solve for a and then substitute k = e^xz/y.

Penny