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Do you remember \[\ln(x^r)=r \ln(x)\]

i got stood up

lol

those loger-epo rules make more sense when defined by calculus

Pretty mkone!

well i think i see something just a little off on your solution

\[4 \ln(3)-2 \ln(4)=\ln(3^4)-\ln(4^2)=\ln(\frac{3^4}{4^2})\]