Solve the following equations: 8−79=67.65 (ii) log, (3x-1)-log, (1−7x)=2 (iii) 31−50(7)+7=0

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

$$x = \frac{\ln\left(\frac{67}{8}\right)}{\ln\left(\frac{7}{6}\right)}$$ $$x = \frac{\ln(67) - \ln(8)}{\ln(7) - \ln(6)}$$ $$x \approx \frac{4.20469 - 2.07944}{1.94591 - 1.79176}$$ $$x \approx \frac{2.12525}{0.15415}$$ $$\boxed{x \approx 13.7869}$$

$$\frac{3x-1}{1-7x} = 3^2$$ $$\frac{3x-1}{1-7x} = 9$$

$$3x-1 = 9(1-7x)$$ $$3x-1 = 9 - 63x$$ $$3x + 63x = 9 + 1$$ $$66x = 10$$ $$x = \frac{10}{66}$$ $$x = \frac{5}{33}$$

For $\log_3(A)$ to be defined, $A > 0$.
For $\log_3(3x-1)$: $3x-1 > 0 \implies 3x > 1 \implies x > \frac{1}{3}$.
For $\log_3(1-7x)$: $1-7x > 0 \implies 1 > 7x \implies x < \frac{1}{7}$.
The value $x = \frac{5}{33} \approx 0.1515$. This value does not satisfy $x > \frac{1}{3}$ (which is $x > 0.3333$) nor does it satisfy $x < \frac{1}{7}$ (which is $x < 0.1428$). In fact, there is no value of $x$ that can satisfy both $x > \frac{1}{3}$ and $x < \frac{1}{7}$ simultaneously.
Therefore, the equation has no solution.

$$\boxed{\text{No solution}}$$

Substitute this into the equation to transform it into a quadratic equation.

$$(7^t)^2 - 50(7^t) + 7 = 0$$ $$y^2 - 50y + 7 = 0$$

Here, $a=1$, $b=-50$, $c=7$.

$$y = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(7)}}{2(1)}$$ $$y = \frac{50 \pm \sqrt{2500 - 28}}{2}$$ $$y = \frac{50 \pm \sqrt{2472}}{2}$$

Simplify the square root: $\sqrt{2472} = \sqrt{4 \times 618} = 2\sqrt{618}$.

$$y = \frac{50 \pm 2\sqrt{618}}{2}$$ $$y = 25 \pm \sqrt{618}$$

Case 1: $7^t = 25 + \sqrt{618}$

$$t = \frac{\ln(25 + \sqrt{618})}{\ln(7)}$$ $$t \approx \frac{\ln(25 + 24.8596)}{\ln(7)}$$ $$t \approx \frac{\ln(49.8596)}{1.9459}$$ $$t \approx \frac{3.9092}{1.9459}$$ $$t_1 \approx 2.0090$$

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

$$t = \frac{\ln(25 - \sqrt{618})}{\ln(7)}$$ $$t \approx \frac{\ln(25 - 24.8596)}{\ln(7)}$$ $$t \approx \frac{\ln(0.1404)}{\ln(7)}$$ $$t \approx \frac{-1.9631}{1.9459}$$ $$t_2 \approx -1.0088$$

Both values of $y$ are positive, so both values of $t$ are valid.

$$\boxed{t \approx 2.0090 \text{ or } t \approx -1.0088}$$