candis
candis
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I see where you're coming from but the answer is no. If you were to graph ln(x) it might be easier to understand. 1: ln is not defined for x <= 0 so it's simply understood that if you declare ln(x) to be a function, x > 0 is implied. 2. Try gr…
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I see where you're coming from but the answer is no. If you were to graph ln(x) it might be easier to understand. 1: ln is not defined for x <= 0 so it's simply understood that if you declare ln(x) to be a function, x > 0 is implied. 2. Try gr…