What is ln(a^{3/2}b^{1/2})-1 ?

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

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simplify ln(a^{3/2}b^{1/2})-1

simplify

ln (a^{\frac{3}{2}} b^{\frac{1}{2}}) - 1

\frac{3}{2} ln (a) + \frac{1}{2} ln (b) - 1

Solution steps

ln (a^{\frac{3}{2}} b^{\frac{1}{2}}) - 1

Apply log rule:

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

ln (a^{\frac{3}{2}} b^{\frac{1}{2}}) = ln (a^{\frac{3}{2}}) + ln (b^{\frac{1}{2}})

= ln (a^{\frac{3}{2}}) + ln (b^{\frac{1}{2}}) - 1

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (a^{\frac{3}{2}}) = \frac{3}{2} ln (a)

= \frac{3}{2} ln (a) + ln (b^{\frac{1}{2}}) - 1

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (b^{\frac{1}{2}}) = \frac{1}{2} ln (b)

= \frac{3}{2} ln (a) + \frac{1}{2} ln (b) - 1

ln (6 e^{x})

lo g_{2} (x) + lo g_{2} (16) + 3 lo g_{2} (\frac{1}{2})

lo g_{7} (\frac{5 z}{6})

3 lo g_{2} (5) + lo g_{2} (7)

4 ln (x) + 2 ln (y) - 3 ln (z)

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