How do you solve ${log}_{2} 3 + {log}_{2} 7 = {log}_{2} x$ $log_2 3 + log_2 7 = log_2 x$ ?

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Answer 1 · 2018-06-08T07:01:09

$x = 21$ $x=21$

Use the addition law of logs.

"If logs are being added, then the numbers are being multiplied"

Note that the bases must be the same for this to apply.

${log}_{a} b + {log}_{a} c = {log}_{a} (b \times c)$ $log_a b + log_a c = log_a(bxxc)$

${log}_{2} 3 + {log}_{2} 7 = {log}_{2} (3 \times 7) = {log}_{2} 21$ $log_2 3+log_2 7 = log_2 (3xx7) = log_2 21$

$x = 21$ $x=21$

How do you solve log23+log27=log2xlog_2 3 + log_2 7 = log_2 x?