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Solve log (small 2)9 + log (small 2)a=log(small 2)13
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\[\log_2 9 + \log_2 a = \log_2 13\] by product law... \[\log_2 9a = \log_2 13\] now i put it in exponential form \[2^{\log_2 13} = 9a\] now according to Logarithmic Property \(a^{\log_a b} = b\) this becomes 13 = 9a a = 13/9
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