<s>[INST] Proving that $f(x) = 6\ln(x^{11}-4) -2$ is one-to-one Please verify my proof, and if there are any mistakes please explain.
Prove that that this function is one-to-one: $f(x) = 6\ln(x^{11}-4) -2$.
Suppose $f(x_1) = f(x_2)$
$\implies 6\ln(x_1^{11}-4) -2  = 6\ln(x_2^{11}-4) -2$
Then we have, $\ln(x_1^{11}-4) - \ln(x_2^{11}-4) = 0$
Using quotient rule for logarithms, $ \ln \frac{x_1^{11}-4}{x_2^{11}-4} = 0\implies e^0 = \frac{x_1^{11}-4}{x_2^{11}-4}$
Thus, $x_1^{11}-4 = x_2^{11}-4 \implies x_1 = x_2$.
So the function  $f(x) = 6ln(x^{11}-4) -2$ is one-to-one
 [/INST] Note that $\log a=\log b\implies a=b $. This reduces two steps.
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