What is 5ln|cot(t)|+5ln(1+tan(2)(t)) ?

The solution to 5ln|cot(t)|+5ln(1+tan(2)(t)) is ln(|cot(t)|^5(1+tan(2)t)^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 5ln|cot(t)|+5ln(1+tan(2)(t))

simplify

5 ln ∣ cot (t) ∣ + 5 ln (1 + tan (2) (t))

ln (∣ cot (t) ∣^{5} (1 + tan (2) t)^{5})

Solution steps

5 ln ∣ cot (t) ∣ + 5 ln (1 + tan (2) (t))

Apply log rule:

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

5 ln ∣ cot (t) ∣ = ln (∣ cot (t) ∣^{5})

= ln (∣ cot (t) ∣^{5}) + 5 ln (tan (2) (t) + 1)

Apply log rule:

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

5 ln (tan (2) (t) + 1) = ln ((tan (2) (t) + 1)^{5})

= ln (∣ cot (t) ∣^{5}) + ln ((tan (2) (t) + 1)^{5})

Apply log rule:

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

ln (∣ cot (t) ∣^{5}) + ln ((1 + tan (2) (t))^{5}) = ln (∣ cot (t) ∣^{5} (tan (2) t + 1)^{5})

= ln (∣ cot (t) ∣^{5} (1 + tan (2) t)^{5})

- lo g_{10} (6.7 \cdot 1 0^{- 4})

ln (2 m^{8})

\frac{1}{11} (lo g_{6} (x) + lo g_{6} (y))

ln (4 x^{2})

lo g_{x} (a) - lo g_{x} (b)

What is 5ln|cot(t)|+5ln(1+tan(2)(t)) ?
The solution to 5ln|cot(t)|+5ln(1+tan(2)(t)) is ln(|cot(t)|^5(1+tan(2)t)^5)