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#.