How do you solve #log _10 (x+21) +log_10 (x)= 2#?

1 Answer
May 18, 2016

See below.

Explanation:

In general: #log_n(a)+log_n(b)=log_n(ab)#
Therefore: #log_10(x+21)+log_10(x)=log_10(x^2+21x)#
By the definition of a logarithm:
#log_10(x^2+21x)=2 => 10^2=x^2+21x#
By solving this as a quadratic:
#100=x^2+21x#
#x^2+21x-100=0#
#(x-4)(x+25)=0#
#x=4# or #x=-25# (are solutions to the equation), but obviously you can't have a negative within a logarithm so x=-25 is not a solution.

It's always important to know the rules of what's being dealt with before attempting it.