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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

simplify

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

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

Solution steps

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

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), assuming x > 0

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

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

Simplify

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

= \frac{4 lo g _{10} ( x ) ( z + 1 )}{y ^{2}}

3 (- 1)^{3} e^{- 1}

[\frac{5}{3}]^{2} 5^{2}

- 1^{2} e^{- 1}

85 \cdot e^{- 862598}

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