How do you use the properties of logarithms to rewrite(contract) each logarithmic expression $2 {log}_{2} (64) + {log}_{2} (2)$ $2log_2(64) + log_2(2)$ ?

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Answer 1 · 2015-07-19T12:59:13

Use ${log}_{a} (a^{b}) = b$ $log_a(a^b) = b$ to find:

$2 {log}_{2} (64) + {log}_{2} (2) = 2 \cdot 6 + 1 = 13$ $2log_2(64)+log_2(2) = 2*6+1 = 13$

${log}_{a} (a^{b}) = b$ $log_a(a^b) = b$ , so

$64 = 2^{6}$ $64=2^6$ so ${log}_{2} (64) = {log}_{2} (2^{6}) = 6$ $log_2(64) = log_2(2^6) = 6$

$2 = 2^{1}$ $2=2^1$ so ${log}_{2} (2) = {log}_{2} (2^{1}) = 1$ $log_2(2) = log_2(2^1) = 1$

So $2 {log}_{2} (64) + {log}_{2} (2) = 2 \cdot 6 + 1 = 13$ $2log_2(64)+log_2(2) = 2*6+1 = 13$

How do you use the properties of logarithms to rewrite(contract) each logarithmic expression 2log_2(64) + log_2(2)2log2(64)+log2(2)2log_2(64) + log_2(2)?