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

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Answer 1 · 2016-09-02T11:33:48

$x approx 10^2/21^10$

$10 log _ 10 (x+21) +log _ 10 (x)= 2$ 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

$x_0 approx 10^2/21^10$

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