What is 1/6 [5ln(x+9)-ln(x)-ln(x^2-4)] ?

The solution to 1/6 [5ln(x+9)-ln(x)-ln(x^2-4)] is 1/6 ln(((x+9)^5)/(x(x^2-4)))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify 1/6 [5ln(x+9)-ln(x)-ln(x^2-4)]

simplify

\frac{1}{6} [5 ln (x + 9) - ln (x) - ln (x^{2} - 4)]

\frac{1}{6} ln (\frac{( x + 9 ) ^{5}}{x ( x ^{2} - 4 )})

Solution steps

\frac{1}{6} [5 ln (x + 9) - ln (x) - ln (x^{2} - 4)]

Apply log rule:

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

5 ln (x + 9) = ln ((x + 9)^{5})

= \frac{1}{6} [ln ((x + 9)^{5}) - ln (x) - ln (x^{2} - 4)]

Apply log rule:

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

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

= \frac{1}{6} [ln (\frac{( x + 9 ) ^{5}}{x}) - ln (x^{2} - 4)]

Apply log rule:

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

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

= \frac{1}{6} [ln (\frac{\frac{( x + 9 ) ^{5}}{x}}{x ^{2} - 4})]

Apply the fraction rule:

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

= \frac{1}{6} [ln (\frac{( x + 9 ) ^{5}}{x ( x ^{2} - 4 )})]

Remove parentheses:

(a) = a

= \frac{1}{6} ln (\frac{( x + 9 ) ^{5}}{x ( x ^{2} - 4 )})

\frac{1}{3} (ln (5) - ln (7))

3.74 + lo g_{10} (\frac{40 + 5 x}{60 - 5 x})

ln (\frac{( 1 - 1.5 )}{0.075}) - 1

3 ln ∣ x ∣ - 4 ln ∣ x + 1 ∣

ln ((4 x^{(2)} - 5) (2 x - 5)^{(3)})

What is 1/6 [5ln(x+9)-ln(x)-ln(x^2-4)] ?
The solution to 1/6 [5ln(x+9)-ln(x)-ln(x^2-4)] is 1/6 ln(((x+9)^5)/(x(x^2-4)))