How do you use the properties of logarithms to expand $ln (\frac{6}{\sqrt{x^{2} + 1}})$ $ln (6/sqrt(x^2+1))$ ?

Question

2764 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-25T15:00:36

$ln 6 - \frac{1}{2} ln (x^{2} + 1)$ $ln6-1/2ln(x^2+1)$

Expand $ln (\frac{6}{\sqrt{x^{2} + 1}})$ $ln(6/(sqrt(x^2+1)))$

Rewrite the square root as an exponent.

$ln (\frac{6}{{(x^{2} + 1)}^{\frac{1}{2}}})$ $ln(6/(x^2+1)^(1/2))$

Use the log property $log (\frac{a}{b}) = log a - log b$ $log(a/b)=loga - logb$

$ln 6 - {ln (x^{2} + 1)}^{\frac{1}{2}}$ $ln6-ln(x^2+1)^(1/2)$

Use the log property ${log a}^{b} = b log a$ $loga^b=bloga$

$ln 6 - \frac{1}{2} ln (x^{2} + 1)$ $ln6-1/2ln(x^2+1)$

How do you use the properties of logarithms to expand ln (6/sqrt(x^2+1))ln(6√x2+1)ln (6/sqrt(x^2+1))?