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log2(a^2-6a)=log2(10+3a)
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a^2-6a = 10+3a solve for a
As the logs have the same base - 2 - we can say that the part in brackets on either side are equivalent i.e. \[a^2 - 6a = 10 +3a\] From there we can solve for a; \[a^2 - 6a - 3a -10 = 0\] \[a^2 - 9a - 10 = 0\] \[(a + 1)(a-10) = 0\] So we can conclude that, \[a = -1\] & \[a = 10\] For this equality to work.
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