How do you solve $log_10a+log_10(a+21)=2$ ?

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Answer 1 · 2016-11-05T13:48:32

$a = 4$ .

Use the rule $log_a(n) + log_a(m) = log_a(n xx m)$ .

$log_10(a(a + 21)) =2$

$a^2 + 21a = 100$

$a^2 + 21a - 100 = 0$

$(a + 25)(a - 4) = 0$

$a = -25 and a = 4$

However, $a= - 25$ is extraneous, since it renders the original equation undefined. The only actual solution is $a = 4$ .

How do you solve log_10a+log_10(a+21)=2log_10a+log_10(a+21)=2?