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)

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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{( 6 + 0 )}{4 - 0})

\frac{4 ln ( 2 )}{ln ( \frac{25}{24} )}

lo g_{e} (\frac{1}{4}) + \frac{1}{4}

\frac{d}{d x} (\frac{3}{4 x}) + ln (2 x^{3})

ln (6) - ln (7) - ln (2) + ln (3)

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)