What is expand log_{3}(1/(x^2y3^z)) ?

The solution to expand log_{3}(1/(x^2y3^z)) is -z-2log_{3}(x)-log_{3}(y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand log_{3}(1/(x^2y3^z))

expand

lo g_{3} (\frac{1}{x ^{2} y 3 ^{z}})

- z - 2 lo g_{3} (x) - lo g_{3} (y)

Solution steps

lo g_{3} (\frac{1}{x ^{2} y \cdot 3 ^{z}})

Apply log rule:

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

lo g_{3} (\frac{1}{x ^{2} y \cdot 3 ^{z}}) = lo g_{3} (1) - lo g_{3} (x^{2} y \cdot 3^{z})

= lo g_{3} (1) - lo g_{3} (3^{z} x^{2} y)

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

lo g_{3} (x^{2} y \cdot 3^{z}) = lo g_{3} (x^{2}) + lo g_{3} (y) + lo g_{3} (3^{z})

= lo g_{3} (1) - (lo g_{3} (x^{2}) + lo g_{3} (y) + lo g_{3} (3^{z}))

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

lo g_{3} (x^{2}) = 2 lo g_{3} (x)

= lo g_{3} (1) - (2 lo g_{3} (x) + lo g_{3} (y) + lo g_{3} (3^{z}))

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

lo g_{3} (3^{z}) = z lo g_{3} (3)

= lo g_{3} (1) - (2 lo g_{3} (x) + lo g_{3} (y) + z lo g_{3} (3))

Apply log rule:

lo g_{a} (1) = 0

= 0 - (lo g_{3} (3) z + 2 lo g_{3} (x) + lo g_{3} (y))

z lo g_{3} (3) = z

= 0 - (z + 2 lo g_{3} (x) + lo g_{3} (y))

0 - (2 lo g_{3} (x) + lo g_{3} (y) + z) = - (2 lo g_{3} (x) + lo g_{3} (y) + z)

= - (z + 2 lo g_{3} (x) + lo g_{3} (y))

Distribute parentheses

= - (z) - (2 lo g_{3} (x)) - (lo g_{3} (y))

Apply minus-plus rules

+ (- a) = - a

= - z - 2 lo g_{3} (x) - lo g_{3} (y)

ln (\frac{273}{259})

4 lo g_{b} (y) + 3 lo g_{b} (x)

2 ln (x) - 4 ln (\frac{1}{y}) - 3 ln (x y)

- 3 ln (2) + ln (3) + 3

ln (\frac{c}{ab})

What is expand log_{3}(1/(x^2y3^z)) ?
The solution to expand log_{3}(1/(x^2y3^z)) is -z-2log_{3}(x)-log_{3}(y)