How do you solve # 10 log _ 10 (x+21) +log _ 10 (x)= 2#?

1 Answer
Cesareo R.
Sep 2, 2016

#x approx 10^2/21^10#

Explanation:

# 10 log _ 10 (x+21) +log _ 10 (x)= 2# or

#(x+21)^(10)x=10^2#

Supposing #x = delta > 0# but very small, we will have

#(delta + 21)^10 approx 21^10# so #21^10 delta approx 10^2#

#delta approx 10^2/21^10# then the real solution is

#x_0 approx 10^2/21^10#

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