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ln(x^2-x)=ln6 use 1-1 property to solve for x ln(x^2-x)=ln6 use 1-1 property to solve for x @Mathematics
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What is this 1-1 property? I can easily solve for x, but if I don't do it your way, your teacher will count it wrong.
the 1-1 property states thta if log base a^x= log base a^y, then x=y
What do you think of this?\[\ln(6)=\ln[(-2)^2-(-2)]\]
yep thts one of the answers. thanks!!
There is a theorem such as\[\log(u)=\log(v) \iff u=v\]which means we can equate the arguments of the logs: ln(x^2-x)=ln6 x^2-x=6 by the theorem x^2-x-6=0 set equal to 0 (x-3)(x+2)=0 factor x=3 or x=-2 solve
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