To solve the equation \(\log x + \log (x - 1) = \log 90\), follow these steps:
1. Combine the logarithmic terms on the left side of the equation:
Recall the property of logarithms that states \(\log a + \log b = \log (a \cdot b)\). Using this property, we can combine the terms on the left side:
[tex]\[
\log x + \log (x - 1) = \log (x \cdot (x - 1))
\][/tex]
2. Rewrite the equation:
Substitute \(\log (x \cdot (x - 1))\) back into the equation:
[tex]\[
\log (x \cdot (x - 1)) = \log 90
\][/tex]
3. Equate the arguments of the logarithms:
Since the logarithm function is one-to-one, if \(\log A = \log B\), then \(A = B\). Therefore, we can write:
[tex]\[
x \cdot (x - 1) = 90
\][/tex]
4. Solve the resulting quadratic equation:
Expand and set the equation to zero:
[tex]\[
x^2 - x - 90 = 0
\][/tex]
This is a standard quadratic equation of the form \(ax^2 + bx + c = 0\).
5. Factor the quadratic equation:
Find two numbers that multiply to \(-90\) and add to \(-1\). These numbers are \(-10\) and \(9\). Thus, we can factor the quadratic equation as:
[tex]\[
(x - 10)(x + 9) = 0
\][/tex]
6. Solve for the roots:
Set each factor to zero and solve for \(x\):
[tex]\[
x - 10 = 0 \quad \text{or} \quad x + 9 = 0
\][/tex]
[tex]\[
x = 10 \quad \text{or} \quad x = -9
\][/tex]
7. Check for extraneous solutions:
Since \( \log x \) and \( \log(x-1) \) are only defined for positive values of \(x\) and \(x-1 > 0\), we need \(x > 1\). Therefore, the solution \(x = -9\) is not valid because it does not satisfy the domain constraint.
The valid solution is:
[tex]\[
x = 10
\][/tex]
Therefore, the solution to the equation \(\log x + \log (x - 1) = \log 90\) is:
[tex]\[
\boxed{10}
\][/tex]
