What is 1/3 ln(x+8)+1/2 ln(x-1) ?

The solution to 1/3 ln(x+8)+1/2 ln(x-1) is ln((x+8)^{1/3}(x-1)^{1/2})

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simplify 1/3 ln(x+8)+1/2 ln(x-1)

simplify

\frac{1}{3} ln (x + 8) + \frac{1}{2} ln (x - 1)

ln ((x + 8)^{\frac{1}{3}} (x - 1)^{\frac{1}{2}})

Solution steps

\frac{1}{3} ln (x + 8) + \frac{1}{2} ln (x - 1)

Apply log rule:

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

\frac{1}{3} ln (x + 8) = ln ((x + 8)^{\frac{1}{3}})

= ln ((x + 8)^{\frac{1}{3}}) + \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 ((x + 8)^{\frac{1}{3}}) + ln ((x - 1)^{\frac{1}{2}})

Apply log rule:

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

ln ((x + 8)^{\frac{1}{3}}) + ln ((x - 1)^{\frac{1}{2}}) = ln ((x + 8)^{\frac{1}{3}} (x - 1)^{\frac{1}{2}})

= ln ((x + 8)^{\frac{1}{3}} (x - 1)^{\frac{1}{2}})

lo g_{a} (x) + lo g_{a} (y) - lo g_{a} (f)

\frac{1}{4} lo g_{b} (x) + 5 lo g_{b} (y) - 3 lo g_{b} (x)

lo g_{a} (\frac{a x}{y ^{5} z ^{8}})

lo g_{10} (x^{4}) y^{5}

What is 1/3 ln(x+8)+1/2 ln(x-1) ?
The solution to 1/3 ln(x+8)+1/2 ln(x-1) is ln((x+8)^{1/3}(x-1)^{1/2})