What is expand ln(\sqrt[5]{(2x-1)/(4x^2+1)}) ?

The solution to expand ln(\sqrt[5]{(2x-1)/(4x^2+1)}) is 1/5 ln((2x-1)/(4x^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ln(\sqrt[5]{(2x-1)/(4x^2+1)})

expand

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

Solution steps

Rewrite as

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

Apply log rule

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

assuming

x \geq 0

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

(8 x^{3} - \frac{12}{x ^{3}}) (4 x^{2} - \frac{6}{x ^{4}})

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

x (\frac{3 x}{4} + \frac{10 - 3 x}{2})

3 (2 x^{3} + 1) (x^{4} + 2 x)^{\frac{1}{2}}

(X^{4} - 2 X^{2} + 2) l e f t (X^{2} - 2 X + 3)

What is expand ln(\sqrt[5]{(2x-1)/(4x^2+1)}) ?
The solution to expand ln(\sqrt[5]{(2x-1)/(4x^2+1)}) is 1/5 ln((2x-1)/(4x^2+1))