## Mathtard23 2 years ago ln(√(6m^4n^2)) write as a sum and/or difference of logarithms, with all variables to the first degree... I know the answer is 1/2ln(6) + 2ln(m) + ln(n) but I don't understand where 1/2 half comes from??

1. ZeHanz

It's the square root.$\ln(\sqrt{6m^4n^2})=\ln(6m^4n^2)^{\frac{1}{2}}=\frac{1}{2}\ln(6m^4n^2)=...$

2. ZeHanz

The rule $lna^b=b \cdot \ln a$ has been used

3. Mathtard23

is the 1/2 used to get rid of the √ (square root)?

4. ZeHanz

√a=a^(1/2), because:$(\sqrt{a})^2=a$and$(a^{\frac{1}{2}})^2=a^{2 \cdot\frac{1}{2}}=a^1=a$ It is just another way of writing √a.

5. Mathtard23

Oooh. I remember that now. Wow you have been a HUGE help. Thank you so much

6. ZeHanz

yw!