How do you simplify $h (x) = log (x^{2} + 6 x + 9) (\sqrt{x^{2} - 1})$ $h(x)= log (x^2+6x+9)(sqrt(x^2-1))$ ?

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Answer 1 · 2018-04-05T11:17:47

$2 log (x + 3) + \frac{1}{2} log (x^{2} - 1)$ $2log(x+3)+1/2log(x^2-1)$

Factor: $x^{2} + 6 x + 9$ $x^2+6x+9$

${(x + 3)}^{2}$ $(x+3)^2$

$\sqrt{x^{2} - 1} = {(x^{2} - 1)}^{\frac{1}{2}}$ $sqrt(x^2-1)=(x^2-1)^(1/2)$

$:.$

$log(x+3)^2(x^2-1)^(1/2)$

$->log(x+3)^2+log(x^2-1)^(1/2)$

$->2log(x+3)+1/2log(x^2-1)$

This is as far as I can go.

How do you simplify h(x)= log (x^2+6x+9)(sqrt(x^2-1))h(x)=log(x2+6x+9)(√x2−1)h(x)= log (x^2+6x+9)(sqrt(x^2-1))?