Exemplary Solutions
Problem 1: Solve for $x$: $\log (x + 5) + \log (x - 5) = \log 24$.
Solution:
Apply the product law $\log A + \log B = \log(AB)$ on the left-hand side:
$$\log[(x + 5)(x - 5)] = \log 24$$
$$\log(x^2 - 25) = \log 24$$
Equate the arguments since logs have the same base:
$$x^2 - 25 = 24 \implies x^2 = 49 \implies x = \pm 7$$
Check Domain Restrictions:
- For $x = 7$: $x + 5 = 12 > 0$ and $x - 5 = 2 > 0$. Valid!
- For $x = -7$: $x - 5 = -12 < 0$. Log of a negative number is undefined! So $x = -7$ is an extraneous root and must be rejected.
$$\mathbf{\text{Unique Valid Solution: } x = 7}$$
Problem 2: If $\log\left(\frac{x+y}{3}\right) = \frac{1}{2}(\log x + \log y)$, prove that $\frac{x}{y} + \frac{y}{x} = 7$.
Solution:
Multiply both sides by $2$:
$$2 \log\left(\frac{x+y}{3}\right) = \log x + \log y$$
Apply power rule on LHS and product rule on RHS:
$$\log\left(\frac{x+y}{3}\right)^2 = \log(xy)$$
Equate the arguments:
$$\left(\frac{x+y}{3}\right)^2 = xy \implies \frac{(x+y)^2}{9} = xy$$
$$(x+y)^2 = 9xy \implies x^2 + 2xy + y^2 = 9xy$$
$$x^2 + y^2 = 7xy$$
Divide both sides by $xy$:
$$\frac{x^2}{xy} + \frac{y^2}{xy} = \frac{7xy}{xy} \implies \mathbf{\frac{x}{y} + \frac{y}{x} = 7} \quad \blacksquare$$
