Question

log2(a^2-6a)=log2(10+3a)

Answer 1

you can cancel out the log_2 if i'm not mistaken since both sides have them

Answer 2

\[Log2(a ^{2}-6a)=Log2(10+3a)\]
\[Log2/Log2=(10+3a)/(a ^{2}-6a)\]
\[1=(10+3a)/(a ^{2}-6a)\]
\[a ^{2}-6a=10+3a\]
\[a ^{2}-6a-3a-10=0\]
\[a ^{2}-9a-10=0\]
\[a ^{2}+a-10a-10=0\]
\[a(a+1)-10(a+1)=0\]
\[(a+1)(a-10)=0\]
\[a+1=0\]
\[a=-1\]
Or,
\[a-10=0\]
\[a=10\]
Therefore, 10 and -1 are the roots of the solution.

Answer 3

if the bases are the same and the exponents (logs) are the same, then the numbers must be the same number. So:
\[a^2-6a=10+3a\]
And
\[a^2-9a-10=0\]
Then proceed as Vengeance has shown