What is 4log_{2}(x)-1/3 log_{2}(x^2+1) ?

The solution to 4log_{2}(x)-1/3 log_{2}(x^2+1) is log_{2}((x^4(x^2+1)^{2/3})/(x^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 4log_{2}(x)-1/3 log_{2}(x^2+1)

simplify

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

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

Solution steps

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

Apply log rule:

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

4 lo g_{2} (x) = lo g_{2} (x^{4})

= lo g_{2} (x^{4}) - \frac{1}{3} lo g_{2} (x^{2} + 1)

Apply log rule:

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

\frac{1}{3} lo g_{2} (x^{2} + 1) = lo g_{2} ((x^{2} + 1)^{\frac{1}{3}})

= lo g_{2} (x^{4}) - lo g_{2} ((x^{2} + 1)^{\frac{1}{3}})

Apply log rule:

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

lo g_{2} (x^{4}) - lo g_{2} ((x^{2} + 1)^{\frac{1}{3}}) = lo g_{2} (\frac{x ^{4}}{( x ^{2} + 1 ) ^{\frac{1}{3}}})

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

Simplify

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

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

ln (ln ((x + 1)^{c o s (4 x)}))

ln (x 1 + x^{2})

f (y) = lo g_{10} (\frac{7 y}{9})

f (b) = lo g_{5} (25 b)

What is 4log_{2}(x)-1/3 log_{2}(x^2+1) ?
The solution to 4log_{2}(x)-1/3 log_{2}(x^2+1) is log_{2}((x^4(x^2+1)^{2/3})/(x^2+1))