Ответ:
1. 81
2. 16
Объяснение:
1. - формула перехода к новому основанию логарифма, =>
- основное логарифмическое тождество
......
3*9*27*....*2187=?
3; 9; 27; ....; 2187 - геометрическая прогрессия
b₁=3, b₂=9, b₃=27, b n=2187. n=?
q=3
n=7
аналогично:
2.
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Verified answer
Ответ:
1. 81
2. 16
Объяснение:
1.
- формула перехода к новому основанию логарифма, =>
......
3*9*27*....*2187=?
3; 9; 27; ....; 2187 - геометрическая прогрессия
b₁=3, b₂=9, b₃=27, b n=2187. n=?
q=3
n=7
аналогично:
2.![\sqrt[9]{3^{\frac{1}{log_{3}2}}*3^{\frac{1}{log_{4}3}}*3^{\frac{1}{log_{8}3}}*...*3^{\frac{1}{log_{256}3}}}=......\sqrt[9]{2*4*8*...*256}=\sqrt[9]{2^{1}*2^{2}*2^{3}*....*2^{8}}=\sqrt[9]{2^{1+2+3+4+5+6+7+8}}=\sqrt[9]{2^{36}}=\sqrt[9]{(2^{4} )^{9}}=2^{4}=16 \sqrt[9]{3^{\frac{1}{log_{3}2}}*3^{\frac{1}{log_{4}3}}*3^{\frac{1}{log_{8}3}}*...*3^{\frac{1}{log_{256}3}}}=......\sqrt[9]{2*4*8*...*256}=\sqrt[9]{2^{1}*2^{2}*2^{3}*....*2^{8}}=\sqrt[9]{2^{1+2+3+4+5+6+7+8}}=\sqrt[9]{2^{36}}=\sqrt[9]{(2^{4} )^{9}}=2^{4}=16](https://tex.z-dn.net/?f=%5Csqrt%5B9%5D%7B3%5E%7B%5Cfrac%7B1%7D%7Blog_%7B3%7D2%7D%7D%2A3%5E%7B%5Cfrac%7B1%7D%7Blog_%7B4%7D3%7D%7D%2A3%5E%7B%5Cfrac%7B1%7D%7Blog_%7B8%7D3%7D%7D%2A...%2A3%5E%7B%5Cfrac%7B1%7D%7Blog_%7B256%7D3%7D%7D%7D%3D......%5Csqrt%5B9%5D%7B2%2A4%2A8%2A...%2A256%7D%3D%5Csqrt%5B9%5D%7B2%5E%7B1%7D%2A2%5E%7B2%7D%2A2%5E%7B3%7D%2A....%2A2%5E%7B8%7D%7D%3D%5Csqrt%5B9%5D%7B2%5E%7B1%2B2%2B3%2B4%2B5%2B6%2B7%2B8%7D%7D%3D%5Csqrt%5B9%5D%7B2%5E%7B36%7D%7D%3D%5Csqrt%5B9%5D%7B%282%5E%7B4%7D%20%29%5E%7B9%7D%7D%3D2%5E%7B4%7D%3D16)