How do you solve ${log}_{10} a + {log}_{10} (a + 21) = 2$ $log_10a+log_10(a+21)=2$ ?

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Answer 1 · 2016-11-05T13:48:32

$a = 4$ $a = 4$ .

Use the rule ${log}_{a} (n) + {log}_{a} (m) = {log}_{a} (n \times m)$ $log_a(n) + log_a(m) = log_a(n xx m)$ .

${log}_{10} (a (a + 21)) = 2$ $log_10(a(a + 21)) =2$

$a^{2} + 21 a = 100$ $a^2 + 21a = 100$

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

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

$a = - 25 and a = 4$ $a = -25 and a = 4$

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

How do you solve log10a+log10(a+21)=2log_10a+log_10(a+21)=2?