Logarithmic Geometric Trigonometry

Given that

$\begin{aligned} a \sin \theta + b \cos \theta & = c & \small \color{#3D99F6} \text{Note that } a = c\cos \theta \text{ and } b = c \sin \theta \\ 2c \sin \theta \cos \theta & = c \\ \sin (2\theta) & = 1 \\ 2 \theta & = 90^\circ \\ \implies \theta & = 45^\circ \end{aligned}$

This means that $\triangle ABC$ is an isosceles right triangle, and $a=b$ and $c = \sqrt 2 a$ . Then we have:

$\begin{aligned} 5 \log_8 a + \frac 43 \log_2 b + \frac 16 \log_{\sqrt 2} c^{12} & = 23 \\ 5 \frac {\log_2 a}{\log_2 2^3} + \frac 43 \log_2 a + 2 \log_{\sqrt 2} (\sqrt 2a) & = 23 \\ 5 \frac {\log_2 a}{\log_2 2^3} + \frac 43 \log_2 a + 2 \left(\log_{\sqrt 2} \sqrt 2+ \log_{\sqrt 2} a \right) & = 23 \\ \frac 53 \log_2 a + \frac 43 \log_2 a + 2 \left(1 + \frac {\log_2a}{\log_2 2^\frac 12} \right) & = 23 \\ 3 \log_2 a + 2 + \frac {2 \log_2a}{\frac 12} & = 23 \\ 7 \log_2 a & = 21 \\ \log_2 a & = 3 \\ a & = 8 \end{aligned}$

Therefore, $k=a+b = 16$ and $(0.25k)^2 - k - 1 = \boxed{-1}$ .

Because $θ$ is angle $BAC$ , $\sin(θ) = \frac{b}{c}$ and $\cos(θ) = \frac{a}{c}$ . By substitution into the trigonometric equation, $2ab = c^2$ . Using the Pythagorean Theorem, $2ab=a^2+b^2$ . This simplifies to $(a-b)^2=0$ by rearranging, factoring, and the Zero-Product property, so $a=b$ . By the Pythagorean Theorem, $c=a√2$ .

Simplifying the logarithmic equation yields $(a^5 * b^4 * c^9 * c^3) = 8^9 * 8^9 * 8^5$ . Simplifying by substitution, $a^9 * a^9 * a^5 = 8^9 * 8^9 * 8^5$ , so $a=b=8$ and $c=8√2$ . By substituting $a+b=a+a=k=16$ into the expression, the final answer is $-1$ .

A trigonometric identity is (for a right triangle) $a\cos \theta + b\sin \theta =c$

Comparing that to $a\sin \theta + b\cos \theta =c$ , we get $\theta = 45°$ , and $a=b$ . Hence, $c=a√2$

Hence, the sum of roots of the equation is $-\frac {\frac {c√2}{2a}}{\frac {b}{a}}=-\frac {bc}{√2 a^2}$ . Putting in the value of c and b, we get the sum of roots as $-1$ .

The logarithmic equation is not needed :)