Can you use the log on question 1

$$\frac{7^x}{6^x} = \frac{67}{8}$$ $$\left(\frac{7}{6}\right)^x = \frac{67}{8}$$

$$x = \frac{\log_{10}(67) - \log_{10}(8)}{\log_{10}(7) - \log_{10}(6)}$$ $$x \approx \frac{1.8260748 - 0.9030900}{0.8450980 - 0.7781513}$$ $$x \approx \frac{0.9229848}{0.0669467}$$ $$\boxed{x \approx 13.7869}$$

This result is identical to the one obtained using natural logarithms, demonstrating that the choice of logarithm base does not affect the final answer.

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We used the property $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$ to combine the two logarithmic terms into a single one:

$$\log_3\left(\frac{3x-1}{1-7x}\right) = 2$$

We converted the logarithmic equation into an exponential equation using the definition $\log_b Y = Z \iff Y = b^Z$:

$$\frac{3x-1}{1-7x} = 3^2$$

This allowed us to eliminate the logarithm and solve for $x$ algebraically.

We checked the domain of the original logarithmic expressions, which requires the arguments of the logarithms to be positive ($3x-1 > 0$ and $1-7x > 0$). This step is crucial for logarithmic equations to ensure that any solutions found are valid. In this case, it led to the conclusion that there was no solution.

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$$t = \frac{\log_{10}(25 + \sqrt{618})}{\log_{10}(7)}$$ $$t \approx \frac{\log_{10}(49.8596)}{\log_{10}(7)}$$ $$t \approx \frac{1.69773}{0.84510}$$ $$t_1 \approx 2.0089$$

Case 2: $7^t = 25 - \sqrt{618}$

$$t = \frac{\log_{10}(25 - \sqrt{618})}{\log_{10}(7)}$$ $$t \approx \frac{\log_{10}(0.1404)}{\log_{10}(7)}$$ $$t \approx \frac{-0.85261}{0.84510}$$ $$t_2 \approx -1.0089$$ $$\boxed{t \approx 2.0089 \text{ or } t \approx -1.0089}$$

Again, the results are consistent with the natural logarithm approach.

The choice of logarithm base (e.g., $\ln$ or $\log_{10}$) for solving exponential equations does not change the final numerical answer, as long as you apply the logarithm consistently to both sides of the equation.