A) \( \Large \frac{1}{log_{3} \pi } + \frac{1}{log_{4} \pi }>2 \) |
B) \( \Large log_{3}5 \) |
C) \( \Large \sqrt{8x} = \frac{10}{3} => x = 16 \) |
D) \( \Large log_{x} \left(a^{2}+1\right)<0 \), (a?0) then 0 |
C) \( \Large \sqrt{8x} = \frac{10}{3} => x = 16 \) |
(A) \( \Large \frac{1}{log_{3} \pi } + \frac{1}{log_{4} \pi } = log_{ \pi }3 + log_{ \pi }4 = log_{ \pi} 12 > 2\)
\( \Large 12 > \pi ^{2} \)
(B) \( \Large log_{3}5 \) is an irrational number
(C) \( \Large log\sqrt{8}x = \frac{10}{3} => x = \left(\sqrt{8}\right)^{\frac{10}{3}} = 2^{5} \)
\( \Large x = 32 \)
(D) \( \Large log_{x} \left(a^{2}+1\right) < 0, a ≠ 0 => a^{2}+1 > 1 \)
So, log is negative
Hence, base is (0, 1).