Question

Need help with this question. ​

Answer 1

We have this equation:

$\log(x) + \log(x + 99) = 2$

First, combine both logarithms using the multiplication property and simplify the expression.

$\log[x(x + 99)] = 2$

$\log[ {x}^{2} + 99x ] = 2$

Now, use the definition of logarithm to transform the equation.

${10}^{2} = {x}^{2} + 99x$

${x}^{2} + 99x - 100 = 0$

Finally, use the quadratic formula to solve the equation.

$x = \frac{ -99 ± \sqrt{ {99}^{2} - 4 \times 1 \times ( - 100)} }{2 \times 1}$

With this, we can say that the solution set is:

• x = 1
• x = -100

We cannot choose x = -100 as a solution because we cannot have a negative logarithm. The only solution is x = 1.