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Log (x+21)+logx=2 solve the equation

2 Answers

7 votes

The sum of logs is the log of the product.

Law of logarithms:


\log_b x + \log_b y = \log_b xy

Apply the law above to the left side of the equation.


\log (x + 21) + \log x = 2


\log [x(x + 21)] = 2


\log (x^2 + 21x) = 2

Now use the definition of log.


\log_b x = y \Leftrightarrow b^y = x


x^2 + 21x = 10^2


x^2 + 21x - 100 = 0


(x + 25)(x - 4) = 0


x + 25 = 0~~~\lor~~~x - 4 = 0


x = -21~~~\lor ~~~x = 4

x = -21 must be discarded because log (x + 21) would become log (-21) which is not defined.

Solution: x = 4

3 votes
x = 4 is the correct answer.
Log (x+21)+logx=2 solve the equation-example-1
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