Show that tanh(.5ln(3)) = 1/2 [working needed]?

Answer 1

tanh is defined as:

tanh(x) = (e^x - e^(-x))/(e^x + e^(-x))

Also by the rules of logs,

.5 ln(3) = ln(3^(.5))

Since e^x and ln(x) are inverse functions,

tanh(.5ln(3)) = (3^(.5) - 3^(-.5))/(3^(.5) + 3^(-.5))

Factoring out 3^(-.5) from top and bottom:

(3^(-.5) (3 - 1))/(3^(-.5) (3 + 1))

Simplifying and canceling:

(3 - 1)/(3 + 1) = 2/4 = 1/2