What is 1+ln(2x)-ln(3y)-2ln(z) ?

The solution to 1+ln(2x)-ln(3y)-2ln(z) is 1+ln((2x)/(3yz^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 1+ln(2x)-ln(3y)-2ln(z)

simplify

1 + ln (2 x) - ln (3 y) - 2 ln (z)

1 + ln (\frac{2 x}{3 y z ^{2}})

Solution steps

1 + ln (2 x) - ln (3 y) - 2 ln (z)

Apply log rule:

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

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

= 1 + ln (\frac{2 x}{3 y}) - 2 ln (z)

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (z) = ln (z^{2})

= 1 + ln (\frac{2 x}{3 y}) - ln (z^{2})

Apply log rule:

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

ln (\frac{2 x}{3 y}) - ln (z^{2}) = ln (\frac{\frac{2 x}{3 y}}{z ^{2}})

= 1 + ln (\frac{\frac{2 x}{3 y}}{z ^{2}})

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= 1 + ln (\frac{2 x}{3 y z ^{2}})

4 lo g_{10} (x) - 2 lo g_{10} (x - 3)

lo g_{2} (r^{\frac{1}{3}}) t

- lo g_{10} (1.9 \cdot 1 0^{- 3})

lo g_{5} (x^{5})

ln (\frac{1}{x ^{2} + y ^{2}})

What is 1+ln(2x)-ln(3y)-2ln(z) ?
The solution to 1+ln(2x)-ln(3y)-2ln(z) is 1+ln((2x)/(3yz^2))