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

1 Answer
Nov 5, 2016

#a = 4#.

Explanation:

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#.