The sum of logs is the log of the product.
Law of logarithms:
[tex] \log_b x + \log_b y = \log_b xy [/tex]
Apply the law above to the left side of the equation.
[tex] \log (x + 21) + \log x = 2 [/tex]
[tex] \log [x(x + 21)] = 2 [/tex]
[tex] \log (x^2 + 21x) = 2 [/tex]
Now use the definition of log.
[tex] \log_b x = y \Leftrightarrow b^y = x [/tex]
[tex] x^2 + 21x = 10^2 [/tex]
[tex] x^2 + 21x - 100 = 0 [/tex]
[tex] (x + 25)(x - 4) = 0 [/tex]
[tex] x + 25 = 0~~~\lor~~~x - 4 = 0 [/tex]
[tex] x = -21~~~\lor ~~~x = 4 [/tex]
x = -21 must be discarded because log (x + 21) would become log (-21) which is not defined.
Solution: x = 4
