How do you solve the equation #log_10z+log_10(z+3)=1#?

1 Answer
Jan 11, 2017

#{2}#

Explanation:

Use the rule #log_a n + log_a m = log_a(n * m)#.

#log_10 z + log_10(z + 3) = 1#

#log_10(z(z + 3)) = 1#

Convert to exponential form.

#z^2 + 3z = 10^1#

#z^2 + 3z = 10#

#z^2 + 3z - 10 = 0#

#(z + 5)(z - 2) = 0#

#z = -5 and 2#

However, #z = -5# is extraneous.

Hopefully this helps!