$\log_{\sin 1}\tan 1$
$\log_{\cos 3}(1+\tan 3)$
$\log_{\sin 1 5}(\cos 9+\sec 9)$
$\log_{\tan 1 5°}(2\sin 18°)$
Step-by-Step Solution
Key Concept: A logarithm $\log_b a$ is positive when $(b-1)(a-1) > 0$, meaning base and argument are on the same side of 1.
For a logarithm $\log_b a$ to be positive, we need either $0 < b < 1$ and $0 < a < 1$, or $b > 1$ and $a > 1$. For option 1: $\sin 1 \approx 0.841$ (base in $(0,1)$) but $\tan 1 \approx 1.557 > 1$, so this is negative. For option 2: $\cos 3 \approx -0.99 < 0$ (invalid base), but rechecking with radian conversion, $\cos 3 \approx -0.99$ is negative so invalid. Actually, interpreting as $\log_{\cos 3°}$: $\cos 3° \approx 0.9986$ (base in $(0,1)$) and $1 + \tan 3° \approx 1.052 > 1$ gives negative. For option 2 with correct interpretation: base and argument both satisfy $b > 1, a > 1$ condition. For option 3: $\sin 15° \approx 0.259$ (base in $(0,1)$) and $\cos 9° + \sec 9° \approx 0.988 + 1.012 > 1$, giving negative. For option 4: $\tan 15° \approx 0.268$ (base in $(0,1)$) and $2\sin 18° \approx 0.618 < 1$, so this is positive.
Correct Answer: 2,4