How do I go about solving this? I know the log rules, I just can't figure it out. Question: log4(x+2) + log4(x-3) = log(4)9 THE "4" IN EACH TERM IS THE BASE. NOT LOG 4, LOG BASE 4.
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Originally Posted by Jeavus How do I go about solving this? I know the log rules, I just can't figure it out. Question: log4(x+2) + log4(x-3) = log(4)9 THE "4" IN EACH TERM IS THE BASE. NOT LOG 4, LOG BASE 4. recall: $\displaystyle \log_a x + \log_a y = \log_a (xy)$ so combine the left hand side using that formula, then equate the things that are being logged on both sides.
Yeah, I know the rules. I've done that, and I'm not getting the correct answer. =/
Originally Posted by Jeavus Yeah, I know the rules. I've done that, and I'm not getting the correct answer. =/ $\displaystyle \log_4 (x + 2) + \log_4 (x - 3) = \log_4 9$ $\displaystyle \Rightarrow \log_4 [(x + 2)(x - 3)] = \log_4 9$ $\displaystyle \Rightarrow (x + 2)(x - 3) = 9$ now solve that quadratic
Ahh, thank you. I didn't think it'd go as far as having to use the quadratic equation.
Originally Posted by Jeavus Ahh, thank you. I didn't think it'd go as far as having to use the quadratic equation. i'm curious as to what you did after you apply the rule. but never mind
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