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How would you solve this: log2 (a^2 -6a)= log2 (10+3a)?
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\[\log_2{(a^2-6a)}=\log_2{(10+3a)} \]
a^2 -6a = 10+3a
a^2 -6 = 10 + 3a a^2 -9a -10 = 0 => a = 10 ( discard a = -1 not belongs to domain of log)
Since the logs and bases are the same, drop them and just solve the two sided equation.
@Chlorophyll how did you get a=10 from a^2 -9a -10=0?
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factor it
You just got answered from lisapeajac!
Oh, see, I'm so used to not skipping steps and showing my work that I got confused when you didn't. But I get it.
That's how you learn :)
Yep. CPR. Get Comfortable with Confusion (C), Persistent in Problemsolving (P), and Reliable in ...Something I forgot... (R). CPR.
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