Question

solve the following logarithmic equation. use a calculator if appropriate. \\(\log x + \log (x + 99) = 2\\) select the correct choice below and, if necessary, fill in the answer box to complete your choice. \\(\bigcirc\\) a. \\(x = \square\\) (simplify your answer. use a comma to separate answers as needed.) \\(\bigcirc\\) b. the solution is the empty set, \\(\varnothing\\).

Explanation:

Step1: Apply log product rule

Using the property \(\log a+\log b = \log(ab)\), we rewrite the equation \(\log x+\log(x + 99)=2\) as \(\log(x(x + 99))=2\). Assuming the base of the logarithm is 10 (since it's not specified, common logarithm base 10 is used), we can convert the logarithmic equation to its exponential form. The exponential form of \(\log_{10}y=k\) is \(y = 10^{k}\). So, \(x(x + 99)=10^{2}=100\).

Step2: Solve the quadratic equation

Expand the left - hand side: \(x^{2}+99x - 100 = 0\).
Factor the quadratic equation: We need two numbers that multiply to \(- 100\) and add up to \(99\). The numbers are \(100\) and \(-1\). So, \(x^{2}+99x - 100=(x + 100)(x - 1)=0\).
Set each factor equal to zero: \(x+100 = 0\) or \(x - 1=0\).
Solving for \(x\), we get \(x=-100\) or \(x = 1\).

Step3: Check the domain of the logarithm

The argument of a logarithm must be positive. For \(\log x\), \(x>0\), and for \(\log(x + 99)\), \(x+99>0\) (i.e., \(x>-99\)).

• When \(x=-100\), \(\log(-100)\) is undefined (since the argument of a logarithm can't be negative).
• When \(x = 1\), \(\log(1)=0\) and \(\log(1 + 99)=\log(100) = 2\), and \(0 + 2=2\), which satisfies the original equation.

Answer:

A. \(x = 1\)