What is 4[ln(x)-ln(x+3)-ln(x-3)] ?

The solution to 4[ln(x)-ln(x+3)-ln(x-3)] is 4ln(x/((x+3)(x-3)))

Popular >

simplify 4[ln(x)-ln(x+3)-ln(x-3)]

simplify

4 [ln (x) - ln (x + 3) - ln (x - 3)]

4 ln (\frac{x}{( x + 3 ) ( x - 3 )})

Solution steps

4 [ln (x) - ln (x + 3) - ln (x - 3)]

Apply log rule:

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

ln (x) - ln (x + 3) = ln (\frac{x}{x + 3})

= 4 [ln (\frac{x}{x + 3}) - ln (x - 3)]

Apply log rule:

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

ln (\frac{x}{x + 3}) - ln (x - 3) = ln (\frac{\frac{x}{x + 3}}{x - 3})

= 4 [ln (\frac{\frac{x}{x + 3}}{x - 3})]

Apply the fraction rule:

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

= 4 [ln (\frac{x}{( x + 3 ) ( x - 3 )})]

Remove parentheses:

(a) = a

= 4 ln (\frac{x}{( x + 3 ) ( x - 3 )})

lo g_{3} (2 x) - lo g_{3} (x)

lo g_{x} (2 \cdot 7)

ln (\frac{z ^{3}}{x y})

lo g_{10} (3 z) + 2 lo g_{10} (7)

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

What is 4[ln(x)-ln(x+3)-ln(x-3)] ?
The solution to 4[ln(x)-ln(x+3)-ln(x-3)] is 4ln(x/((x+3)(x-3)))