What is expand log_{10}(x^4)y^{-2}(z+1) ?

The solution to expand log_{10}(x^4)y^{-2}(z+1) is (4zlog_{10}(x)+4log_{10}(x))/(y^2)

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expand log_{10}(x^4)y^{-2}(z+1)

expand

lo g_{10} (x^{4}) y^{- 2} (z + 1)

\frac{4 z lo g _{10} ( x ) + 4 lo g _{10} ( x )}{y ^{2}}

Solution steps

lo g_{10} (x^{4}) y^{- 2} (z + 1)

= y^{- 2} lo g_{10} (x^{4}) (z + 1)

Apply the distributive law:

a (b + c) = ab + a c

y^{- 2} lo g_{10} (x^{4}) (z + 1) = y^{- 2} lo g_{10} (x^{4}) z + y^{- 2} lo g_{10} (x^{4}) \cdot 1

= y^{- 2} lo g_{10} (x^{4}) z + y^{- 2} lo g_{10} (x^{4}) \cdot 1

Simplify

y^{- 2} lo g_{10} (x^{4}) z + y^{- 2} lo g_{10} (x^{4}) \cdot 1 : \frac{4 z lo g _{10} ( x ) + 4 lo g _{10} ( x )}{y ^{2}}

= \frac{4 z lo g _{10} ( x ) + 4 lo g _{10} ( x )}{y ^{2}}

det (8 + j^{8}) (j^{2}) (j^{2}) (4 - j^{4})

x + 3 x + 6

(e^{- 2 x}) (e^{x})

440 \cdot 10 \cdot 1 0^{23} \cdot (π \cdot (0.1 \cdot 1 0^{- 3})^{2})

(5 d^{8} e^{2}) (9 d^{- 4} e^{- 2})

What is expand log_{10}(x^4)y^{-2}(z+1) ?
The solution to expand log_{10}(x^4)y^{-2}(z+1) is (4zlog_{10}(x)+4log_{10}(x))/(y^2)