What is 1/4 log_{e}(y-2)-1/4 log_{e}(y+2) ?

The solution to 1/4 log_{e}(y-2)-1/4 log_{e}(y+2) is ln(\sqrt[4]{(y-2)/(y+2)})

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simplify 1/4 log_{e}(y-2)-1/4 log_{e}(y+2)

simplify

\frac{1}{4} lo g_{e} (y - 2) - \frac{1}{4} lo g_{e} (y + 2)

Solution steps

\frac{1}{4} lo g_{e} (y - 2) - \frac{1}{4} lo g_{e} (y + 2)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{4} lo g_{e} (y - 2) = lo g_{e} ((y - 2)^{\frac{1}{4}})

= lo g_{e} ((y - 2)^{\frac{1}{4}}) - \frac{1}{4} lo g_{e} (y + 2)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{4} lo g_{e} (y + 2) = lo g_{e} ((y + 2)^{\frac{1}{4}})

= lo g_{e} ((y - 2)^{\frac{1}{4}}) - lo g_{e} ((y + 2)^{\frac{1}{4}})

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

lo g_{e} ((y - 2)^{\frac{1}{4}}) - lo g_{e} ((y + 2)^{\frac{1}{4}}) = lo g_{e} (\frac{( y - 2 ) ^{\frac{1}{4}}}{( y + 2 ) ^{\frac{1}{4}}})

= lo g_{e} (\frac{( y - 2 ) ^{\frac{1}{4}}}{( y + 2 ) ^{\frac{1}{4}}})

Combine same powers :

Apply log rule:

lo g_{e} (x) = ln (x)

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What is 1/4 log_{e}(y-2)-1/4 log_{e}(y+2) ?
The solution to 1/4 log_{e}(y-2)-1/4 log_{e}(y+2) is ln(\sqrt[4]{(y-2)/(y+2)})