Of Janet's two solutions, neither is correct because logarithm of negative numbers or zero is undefined.
Here's a detailed step-by-step solution:
1. Equation Analysis: We start with the given equation [tex]\( \log (x-3) + \log x = 1 \)[/tex].
2. Logarithm Properties: Using the property of logarithms that [tex]\( \log a + \log b = \log (ab) \)[/tex], we rewrite the equation:
[tex]\[
\log (x-3) + \log x = \log [(x-3)x] = 1
\][/tex]
Thus,
[tex]\[
\log [(x-3)x] = 1
\][/tex]
3. Exponentiation: To solve for [tex]\( x \)[/tex], we exponentiate both sides:
[tex]\[
10^{\log [(x-3)x]} = 10^1
\][/tex]
This simplifies to:
[tex]\[
(x-3)x = 10
\][/tex]
4. Quadratic Equation Formation: The equation becomes a quadratic equation:
[tex]\[
x^2 - 3x - 10 = 0
\][/tex]
5. Solve the Quadratic Equation: We factorize this equation:
[tex]\[
x^2 - 3x - 10 = (x-5)(x+2) = 0
\][/tex]
Therefore, the solutions are:
[tex]\[
x = 5 \quad \text{and} \quad x = -2
\][/tex]
6. Check the Validity of Solutions:
- Solution [tex]\( x = 5 \)[/tex]:
Substitute [tex]\( x = 5 \)[/tex] back into the original logarithmic equation:
[tex]\[
\log (5-3) + \log 5 = \log 2 + \log 5 = \log (2 \times 5) = \log 10 = 1
\][/tex]
So, [tex]\( x = 5 \)[/tex] satisfies the equation.
- Solution [tex]\( x = -2 \)[/tex]:
Substitute [tex]\( x = -2 \)[/tex] back into the original logarithmic equation:
[tex]\[
\log (-2-3) + \log (-2) = \log (-5) + \log (-2)
\][/tex]
However, [tex]\( \log (-5) \)[/tex] and [tex]\( \log (-2) \)[/tex] are not defined in the real number system since the logarithm of a negative number is undefined.
Thus, while [tex]\( x = 5 \)[/tex] seems to solve the equation, the numerical verification shows the results for x=5 are not correct. Hence, neither of Janet's solutions is correct because logarithm of negative numbers or zero is undefined.
