How do you expand $ln \sqrt{(x^{2} (x + 2))}$ $ln sqrt (( x^2( x+2))$ ?

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How do you expand $ln \sqrt{(x^{2} (x + 2))}$ $ln sqrt (( x^2( x+2))$ ?

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Answer 1 · 2016-05-30T21:33:32

For this problem, you must remember that $\sqrt{x} = x^{\frac{1}{2}}$ $sqrt(x) = x^(1/2)$

${ln (x^{2} (x + 2))}^{\frac{1}{2}}$ $ln(x^2(x + 2))^(1/2)$

Using the product rule ( ${log}_{a} n + {log}_{a} m = {log}_{a} (n \times m)$ $log_an + log_am = log_a(n xx m)$ ) and distributing.

$= {ln (x^{2})}^{\frac{1}{2}} + {ln (x + 2)}^{\frac{1}{2}}$ $=ln(x^2)^(1/2) + ln(x + 2)^(1/2)$

Using exponent rules to simplify $({(x^{2})}^{\frac{1}{2}} = x^{2 \times \frac{1}{2}} = x^{1}$ $((x^2)^(1/2) =x^ (2 xx 1/2) = x^1$ )
$= ln x + {ln (x + 2)}^{\frac{1}{2}}$ $=lnx + ln(x + 2)^(1/2)$

Now use the log rule ${log n}^{a} = a log n$ $logn^a = alogn$

$= ln x + \frac{1}{2} ln (x + 2)$ $=lnx + 1/2ln(x + 2)$

Hopefully this helps!

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How do you expand $ln \sqrt{(x^{2} (x + 2))}$ $ln sqrt (( x^2( x+2))$ ?

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