Isn't the natural log of 1 equal to zero? If so, doesn't that make this method incorrect?

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You are right, she made a mistake but the answer comes out the same. The natural log of 1 is zero, but ln(A*B)=ln(A)+ln(B) so you just add zero to the natural log of (1/2)^(t/5730). The easier way to think about it would be to divide both sides by 1 before you take the natural log so it goes away. Or to just realize that (1)*(1/2)^(t/5370) is equal to (1/2)^(t/5370). Hope that helps :D