What is ln(x)-ln(x-4)+1/2 ln(x+1) ?

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

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simplify ln(x)-ln(x-4)+1/2 ln(x+1)

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

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

Multiply fractions:

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

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

\frac{1}{2} ln (x) + 6 ln (y) - 9 ln (z)

lo g_{10} (\frac{- 5 ^{7} 2 ^{5}}{- 1 0 ^{8}})

ln (3 e^{7 x^{2} - 13 x + 5})

lo g_{b} (u^{5} v^{2})

- ln (x) + ln (c)

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