Question

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

Answer 1

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.