What is 2ln(2)+9/5 ln(4)+1/5 ln(-1) ?

The solution to 2ln(2)+9/5 ln(4)+1/5 ln(-1) is ln(-32*2^{3/5})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 2ln(2)+9/5 ln(4)+1/5 ln(-1)

simplify

2 ln (2) + \frac{9}{5} ln (4) + \frac{1}{5} ln (- 1)

ln (- 32 \cdot 2^{\frac{3}{5}})

Solution steps

2 ln (2) + \frac{9}{5} ln (4) + \frac{1}{5} ln (- 1)

Apply log rule:

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

\frac{9}{5} ln (4) = ln (4^{\frac{9}{5}})

= 2 ln (2) + ln (4^{\frac{9}{5}}) + \frac{1}{5} ln (- 1)

Apply log rule:

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

\frac{1}{5} ln (- 1) = ln ((- 1)^{\frac{1}{5}})

= 2 ln (2) + ln (4^{\frac{9}{5}}) + ln ((- 1)^{\frac{1}{5}})

Apply log rule:

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

ln (4^{\frac{9}{5}}) + ln ((- 1)^{\frac{1}{5}}) = ln (4^{\frac{9}{5}} (- 1)^{\frac{1}{5}})

= 2 ln (2) + ln (4^{\frac{9}{5}} (- 1)^{\frac{1}{5}})

4^{\frac{9}{5}} (- 1)^{\frac{1}{5}} = - 4 \cdot 4^{\frac{4}{5}}

= 2 ln (2) + ln (- 4 \cdot 4^{\frac{4}{5}})

Apply log rule:

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

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

= ln (2^{2}) + ln (- 4 \cdot 4^{\frac{4}{5}})

Apply log rule:

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

ln (2^{2}) + ln (- 4 \cdot 4^{\frac{4}{5}}) = ln (2^{2} (- 4 \cdot 4^{\frac{4}{5}}))

= ln (2^{2} (- 4 \cdot 4^{\frac{4}{5}}))

2^{2} (- 4 \cdot 4^{\frac{4}{5}}) = - 32 \cdot 2^{\frac{3}{5}}

= ln (- 32 \cdot 2^{\frac{3}{5}})

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

lo g_{5} (x y^{2})

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

ln (1200 - y) + ln (y - 100)

\frac{2 ln ( 3 )}{ln ( 2 ) - ln ( 3 )}

What is 2ln(2)+9/5 ln(4)+1/5 ln(-1) ?
The solution to 2ln(2)+9/5 ln(4)+1/5 ln(-1) is ln(-32*2^{3/5})