Expand this logarithmic equation (using properties) : $ln (\frac{\sqrt{x - 5}}{y^{2}})$ $ln(sqrt(x-5)/y^2)$ how do i do this?

Question

this is what my teacher did but i think she did it wrong and i feel like the $\frac{1}{5}$ $1/5$ is supposed to be $\frac{1}{2}$ $1/2$ . someone pls help and tell me if im wrong or right!

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Answer 1 · 2017-12-21T01:06:24

You are correct that it should be $\frac{1}{2}$ $1/2$ :

$ln (\frac{\sqrt{x - 5}}{y^{2}}) = (\frac{1}{2}) ln (x - 5) - (2) ln (y)$ $ln(sqrt(x-5)/y^2) = (1/2)ln(x-5)-(2)ln(y)$

Given: $ln (\frac{\sqrt{x - 5}}{y^{2}})$ $ln(sqrt(x-5)/y^2)$

Use the property $ln (\frac{a}{b}) = ln (a) - ln (b)$ $ln(a/b) = ln(a) - ln(b)$ where $a = \sqrt{x - 5} and$ $a = sqrt(x-5) and$ b = y^2#:

$ln (\frac{\sqrt{x - 5}}{y^{2}}) = ln (\sqrt{x - 5}) - ln (y^{2})$ $ln(sqrt(x-5)/y^2) = ln(sqrt(x-5))-ln(y^2)$

Write $\sqrt{x - 5}$ $sqrt(x-5)$ as ${(x - 5)}^{\frac{1}{2}}$ $(x-5)^(1/2)$ :

$ln (\frac{\sqrt{x - 5}}{y^{2}}) = ln ({(x - 5)}^{\frac{1}{2}}) - ln (y^{2})$ $ln(sqrt(x-5)/y^2) = ln((x-5)^(1/2))-ln(y^2)$

Use the property $ln (a^{c}) = (c) ln (a)$ $ln(a^c) = (c)ln(a)$ on the first term:

$ln (\frac{\sqrt{x - 5}}{y^{2}}) = (\frac{1}{2}) ln (x - 5) - ln (y^{2})$ $ln(sqrt(x-5)/y^2) = (1/2)ln(x-5)-ln(y^2)$

Do the same on the second term:

$ln (\frac{\sqrt{x - 5}}{y^{2}}) = (\frac{1}{2}) ln (x - 5) - (2) ln (y)$ $ln(sqrt(x-5)/y^2) = (1/2)ln(x-5)-(2)ln(y)$