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\[\log_{6}(a ^{2}+2)+\log_{6}2=2\]
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$$ log_6(a^2+2)+log_6(2)=2\\ \implies log_6\pmatrix{(a^2+2)\times 2} =2 \implies log_6(2a^2+4)=2\\ \implies 6^{log_6(2a^2+4)}=6^2\\ \text{using the log cancellation rules}\\ \implies 2a^2+4 = 6^2 $$
and I think you can get it from there
Adding logarithms is the same as taking the logarithm of the product. You can rewrite that equation as \[\log_6(2(a^2+2)) = 2\]Then raise 6 to the power of each side, and remember that \(b^{\log_b a} = a\) so you have\[2(a^2+2) = 6^2\]
x=4
yes
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and...
x=-4
right.
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