Question: 8 — Solving Logarithmic Equations using Exponents
LEARNING OBJECTIVE: Solve a logarithmic equation by rewriting into an exponential equation.
If $\log_5(-3x + 5) = 3$, then what is the value of $x$?
a.) 125
b.) -40
c.) 50
d.) -25
1. **State the problem:** We need to solve the logarithmic equation $$\log_5(-3x + 5) = 3$$ for $x$.
2. **Recall the definition of logarithm:** The equation $\log_b(a) = c$ is equivalent to the exponential form $$a = b^c$$.
3. **Rewrite the logarithmic equation in exponential form:**
$$-3x + 5 = 5^3$$
4. **Calculate the right side:**
$$5^3 = 125$$
So,
$$-3x + 5 = 125$$
5. **Isolate $x$:**
$$-3x = 125 - 5$$
$$-3x = 120$$
6. **Divide both sides by $-3$:**
$$x = \frac{120}{-3}$$
Show cancellation:
$$x = \frac{\cancel{120}}{\cancel{-3}} = -40$$
7. **Check the domain:** The argument of the logarithm must be positive:
$$-3x + 5 > 0$$
$$-3x > -5$$
$$x < \frac{5}{3}$$
Since $x = -40$ satisfies $x < \frac{5}{3}$, it is valid.
**Final answer:** $x = -40$ (option b).