Sets, Relations & Functions Question - None

Question

If $f(x) = \log_e \left( \frac{1 - x}{1 + x} \right) |x| < 1$, then $f \left( \frac{2x}{1 + x^2} \right)$ is equal to :

Accepted Answer

Let u=8^{2x}. y=(u²-1)/(u²+1). Solve: u²=(1+y)/(1-y). So 4x=log₈((1+y)/(1-y))=(log₈e)ln((1+y)/(1-y)). f⁻¹(x)=(log₈e/4)ln((1+x)/(1-x)).

Trap: Remember log₈A=(log₈e)ln A.

Key Concept: (t-1/t)/(t+1/t) = (t²-1)/(t²+1) standard form

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