What is expand (ln^2(x)-ln(x)-2)/((x(ln(x)-4))) ?

The solution to expand (ln^2(x)-ln(x)-2)/((x(ln(x)-4))) is -2/(x(ln(x)-4))

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expand (ln^2(x)-ln(x)-2)/((x(ln(x)-4)))

expand

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

- \frac{2}{x ( ln ( x ) - 4 )}

Solution steps

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

Apply log rule:

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

ln^{2} (x) - ln (x) = ln^{2} (\frac{x}{x})

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

Apply rule

\frac{a}{a} = 1

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

ln^{2} (1) = 0

= \frac{0 - 2}{x ( ln ( x ) - 4 )}

Subtract the numbers:

0 - 2 = - 2

= \frac{- 2}{x ( ln ( x ) - 4 )}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{x ( ln ( x ) - 4 )}

ln (9^{3 x - 2}) + ln (9^{3 x - 3})

ln (- 2) + ln (- 3)

lo g_{10} (t^{x} (1 - t)^{1 - x})

0.48 - \frac{0.0592}{6} lo g_{10} (\frac{1.5}{0.5})

What is expand (ln^2(x)-ln(x)-2)/((x(ln(x)-4))) ?
The solution to expand (ln^2(x)-ln(x)-2)/((x(ln(x)-4))) is -2/(x(ln(x)-4))