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

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The discussion focuses on proving that (fοg)(x) equals 3f(x) for the given functions f(x) and g(x). The proof begins by substituting g(x) into f(x) and simplifying the logarithmic expression. After clearing denominators and rearranging, the logarithmic terms are expressed in terms of cubes, leading to the relationship between f(g(x)) and f(x). The final step applies the logarithmic property to confirm that f(g(x)) equals 3f(x), completing the proof. The discussion emphasizes the importance of algebraic manipulation and logarithmic rules in the proof process.
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f(x)=log(1+x)/(1-x) and g(x)=(3x+x2)/(3x2+1) prove that (fοg)(x)=3f(x)
 
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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?
 
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 :):)
 
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. :)
 
thank you so much :)
 
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