Proving (fοg)(x) = 3f(x) for f(x)=log(1+x)/(1-x) and g(x)=(3x+x2)/(3x2+1)

[spartas]

f(x)=log(1+x)/(1-x) and g(x)=(3x+x²)/(3x²+1) prove that (fοg)(x)=3f(x)

 

[MarkFL]

In order for us to actually provide help, we need to see what you've tried so we know where you are having trouble. :D

We know that:

$$(f\,\circ\,g)(x)\equiv f(g(x))$$

It seems we actually need:

$$f(x)=\log\left(\frac{1+x}{1-x}\right)$$

$$g(x)=\frac{3x+x^3}{3x^2+1}$$

So, in the definition of $f$, wherever there is an $x$ we want to put in the definition of $g$:

$$f(g(x))=\log\left(\frac{1+\dfrac{3x+x^3}{3x^2+1}}{1-\dfrac{3x+x^3}{3x^2+1}}\right)$$

Can you now simplify this using some algebra and a logarithmic rule to get the desired result?

 

[spartas]

im having problems at the logarithm part because I am not good at it! anyway thank you for taking some of your time to answer :):)

 

[MarkFL]

This is how I would finish the problem:

$$f(g(x))=\log\left(\frac{1+\dfrac{3x+x^3}{3x^2+1}}{1-\dfrac{3x+x^3}{3x^2+1}}\right)$$

Multiply the argument of the log function by $$1=\frac{3x^2+1}{3x^2+1}$$ to clear the denominators:

$$f(g(x))=\log\left(\frac{1+\dfrac{3x+x^3}{3x^2+1}}{1-\dfrac{3x+x^3}{3x^2+1}}\cdot\frac{3x^2+1}{3x^2+1}\right)$$

$$f(g(x))=\log\left(\frac{3x^2+1+3x+x^3}{3x^2+1-3x-x^3}\right)$$

Do some rearranging:

$$f(g(x))=\log\left(\frac{1+3x+3x^2+x^3}{1-3x+3x^2-x^3}\right)$$

Now, note that:

$$(1+x)^3=1+3x+3x^2+x^3$$ and $$(1-x)^3=1-3x+3x^2-x^3$$

And so we may write:

$$f(g(x))=\log\left(\frac{(1+x)^3}{(1-x)^3}\right)$$

$$f(g(x))=\log\left(\left(\frac{1+x}{1-x}\right)^3\right)$$

Next, we apply the logarithmic property $$\log_a\left(b\,^c\right)=c\cdot\log_a(b)$$ to obtain:

$$f(g(x))=3\log\left(\frac{1+x}{1-x}\right)=3\cdot f(x)$$

Shown as desired. :)

 

[spartas]

thank you so much :)