If $0 < x < 2\pi$, then the number of real values of $x$ satisfying the equation $8^{\sin x} + 8^{\sin(x-1)} = 30$ is
Step-by-Step Solution
Key Concept: Using substitution to convert exponential equations into polynomial form, then solving and checking validity of solutions against domain constraints
Let $8^{\sin x} = t$. Then $8t^{\sin x} = 81^{\sin x} \cdot \frac{8t}{81}$. So the given equation is $t + \frac{8t}{30} = 81 \Rightarrow t^2 - 30t + 81 = 0 \Rightarrow t = 3$ or $27$. We have $8^{\sin x} = 3$ or $27$, giving $\sin x = \log_8 3$ or $\sin x = 3 \log_8 3$. The value satisfying $-1 \leq \sin x \leq 1$ is $8^{\sin x} = 3$ or $27$.
Correct Answer: 8