What is expand log_{10}(sqrt(x^9y^{-8)}) ?

The solution to expand log_{10}(sqrt(x^9y^{-8)}) is 9/2 log_{10}(x)-4log_{10}(y)

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expand log_{10}(sqrt(x^9y^{-8)})

expand

lo g_{10} (x^{9} y^{- 8})

\frac{9}{2} lo g_{10} (x) - 4 lo g_{10} (y)

Solution steps

lo g_{10} (x^{9} y^{- 8})

x^{9} y^{- 8} = \frac{x ^{4} x}{y ^{4}}

= lo g_{10} (\frac{x ^{4} x}{y ^{4}})

Apply log rule:

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

lo g_{10} (\frac{x ^{4} x}{y ^{4}}) = lo g_{10} (x^{4} x) - lo g_{10} (y^{4})

= lo g_{10} (x^{4} x) - lo g_{10} (y^{4})

Simplify

lo g_{10} (y^{4}) : 4 lo g_{10} (y)

lo g_{10} (x^{4} x) = \frac{9}{2} lo g_{10} (x)

= \frac{9}{2} lo g_{10} (x) - 4 lo g_{10} (y)

e^{- 5 (x + 1)} \cdot δ (x - 3) + cos (5 \cdot π \cdot x) \cdot δ (x - 4)

(x^{3} - \frac{1}{x ^{6}})^{6}

[- 1.128 e^{- 3.975 x} + 5.038]^{2}

\frac{2}{( 3 - 1 ) ^{3}} + sin (3 - 1)

\frac{4 ( 2 x z ^{π} - y t ) ( 2 x z ^{π} + y t )}{y ^{2} z ^{2 π}}

What is expand log_{10}(sqrt(x^9y^{-8)}) ?
The solution to expand log_{10}(sqrt(x^9y^{-8)}) is 9/2 log_{10}(x)-4log_{10}(y)