What is 1/2 log_{10}(2x)-5/2 log_{10}(y) ?

The solution to 1/2 log_{10}(2x)-5/2 log_{10}(y) is log_{10}((sqrt(2)sqrt(x))/(y^{5/2)})

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simplify 1/2 log_{10}(2x)-5/2 log_{10}(y)

simplify

\frac{1}{2} lo g_{10} (2 x) - \frac{5}{2} lo g_{10} (y)

lo g_{10} (\frac{2 x}{y ^{\frac{5}{2}}})

Solution steps

\frac{1}{2} lo g_{10} (2 x) - \frac{5}{2} lo g_{10} (y)

Apply log rule:

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

\frac{1}{2} lo g_{10} (2 x) = lo g_{10} ((2 x)^{\frac{1}{2}})

= lo g_{10} ((2 x)^{\frac{1}{2}}) - \frac{5}{2} lo g_{10} (y)

Apply log rule:

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

\frac{5}{2} lo g_{10} (y) = lo g_{10} (y^{\frac{5}{2}})

= lo g_{10} ((2 x)^{\frac{1}{2}}) - lo g_{10} (y^{\frac{5}{2}})

Apply log rule:

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

lo g_{10} ((2 x)^{\frac{1}{2}}) - lo g_{10} (y^{\frac{5}{2}}) = lo g_{10} (\frac{( 2 x ) ^{\frac{1}{2}}}{y ^{\frac{5}{2}}})

= lo g_{10} (\frac{( 2 x ) ^{\frac{1}{2}}}{y ^{\frac{5}{2}}})

(2 x)^{\frac{1}{2}} = 2 x

= lo g_{10} (\frac{2 x}{y ^{\frac{5}{2}}})

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What is 1/2 log_{10}(2x)-5/2 log_{10}(y) ?
The solution to 1/2 log_{10}(2x)-5/2 log_{10}(y) is log_{10}((sqrt(2)sqrt(x))/(y^{5/2)})