Exponential Logs 917F96

1. **State the problem:** We want to solve the exponential equation $$3^{2x} = 75$$ for $x$. 2. **Recall the logarithm definition:** Logarithms are the inverse operations of exponentials. If $$a^b = c$$, then $$\log_a c = b$$. This means taking the logarithm base $a$ of $c$ gives the exponent $b$. 3. **Apply logarithms to both sides:** To solve for $x$, we take the logarithm base $3$ of both sides because the base of the exponential is $3$. This gives: $$\log_3(3^{2x}) = \log_3(75)$$ 4. **Use the logarithm power rule:** The logarithm of a power is the exponent times the logarithm of the base: $$\log_3(3^{2x}) = 2x \cdot \log_3(3)$$ Since $$\log_3(3) = 1$$ (because $3^1=3$), this simplifies to: $$2x = \log_3(75)$$ 5. **Solve for $x$:** Divide both sides by $2$: $$x = \frac{\log_3(75)}{2}$$ 6. **Is it illicit to write $\log_3 75 = 2x$ and solve?** No, it is not illicit. This is exactly the step we used. Writing $$\log_3 75 = 2x$$ is correct and a standard way to solve exponential equations by taking logarithms with the same base as the exponential. 7. **If you want to calculate $x$ using common logarithms (base 10) or natural logarithms (base $e$):** Use the change of base formula: $$\log_3(75) = \frac{\log(75)}{\log(3)}$$ Then $$x = \frac{1}{2} \cdot \frac{\log(75)}{\log(3)}$$ **Summary:** Taking logarithms allows us to "bring down" the exponent so we can solve for the variable. Using the same base logarithm as the exponential base is the most straightforward method, and writing $$\log_3 75 = 2x$$ is perfectly valid.