Question

solve for x:
log(x) + log(x + 30) = 3
x =
you may enter the exact value or round to 4 decimal places.
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Explanation:

Step1: Apply log product rule

Using the property \(\log(a) + \log(b)=\log(ab)\), we rewrite the equation:
\(\log(x(x + 30))=3\)

Step2: Convert to exponential form

Assuming the logarithm is base 10 (since it's not specified, standard base 10), we convert \(\log_{10}(x(x + 30)) = 3\) to exponential form:
\(x(x + 30)=10^{3}=1000\)

Step3: Expand and form quadratic

Expand the left - hand side: \(x^{2}+30x - 1000 = 0\)

Step4: Solve quadratic equation

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here, \(a = 1\), \(b = 30\), \(c=- 1000\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(30)^{2}-4\times1\times(-1000)=900 + 4000 = 4900\)
Then, \(x=\frac{-30\pm\sqrt{4900}}{2\times1}=\frac{-30\pm70}{2}\)
We have two solutions:
\(x_{1}=\frac{-30 + 70}{2}=\frac{40}{2}=20\)
\(x_{2}=\frac{-30 - 70}{2}=\frac{-100}{2}=-50\)

But we need to check the domain of the logarithmic function. For \(\log(x)\) and \(\log(x + 30)\) to be defined, \(x>0\) and \(x + 30>0\) (i.e., \(x>-30\)). So \(x=-50\) is not in the domain.

Answer:

\(20\)