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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??

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It's the square root.\[\ln(\sqrt{6m^4n^2})=\ln(6m^4n^2)^{\frac{1}{2}}=\frac{1}{2}\ln(6m^4n^2)=...\]
The rule \[lna^b=b \cdot \ln a\] has been used
is the 1/2 used to get rid of the √ (square root)?

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Other answers:

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

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