Can you solve this logartihm problem?
回答 (3)
2017-10-30 6:13 am
Use log_x(a)/log_x(b) = log_b(a)
3/log_2(16) + 2/log_5(16) - 2/log_10(16)
= log_2(8)/log_2(16) + log_5(25)/log_5(16) - log_10(100)/log_10(16)
= log_16(8) + log_16(25) - log_16(100)
= log_16(8*25/100)
= log_16(2)
= log_16(16^1/4)
= 1/4
3/log_2(16) + 2/log_5(16) - 2/log_10(16)
= log_2(8)/log_2(16) + log_5(25)/log_5(16) - log_10(100)/log_10(16)
= log_16(8) + log_16(25) - log_16(100)
= log_16(8*25/100)
= log_16(2)
= log_16(16^1/4)
= 1/4
2017-10-30 8:16 pm
I get 1/log 4. Is it correct?
2017-10-30 9:16 am
Apply the change of base formula:
log[b](a) = log[c](a)/log[c](b)
= log(a)/log(b)
1/log[b](a) = log(b)/log(a)
= log[a](b)
3/log₂16 + 2/log₅16 -2/log₁₀16
= 3log₁₆2 + 2log₁₆5 -2log₁₆10
= log₁₆(2³)+log₁₆(5²)-log₁₆(10²)
= log₁₆(8)+log₁₆(25)-log₁₆(100)
= log₁₆(8×25/100)
= log₁₆(200/100)
= log₁₆(2) = 1/4
// 16^(1/4) = 2
log[b](a) = log[c](a)/log[c](b)
= log(a)/log(b)
1/log[b](a) = log(b)/log(a)
= log[a](b)
3/log₂16 + 2/log₅16 -2/log₁₀16
= 3log₁₆2 + 2log₁₆5 -2log₁₆10
= log₁₆(2³)+log₁₆(5²)-log₁₆(10²)
= log₁₆(8)+log₁₆(25)-log₁₆(100)
= log₁₆(8×25/100)
= log₁₆(200/100)
= log₁₆(2) = 1/4
// 16^(1/4) = 2
收錄日期: 2021-05-01 21:54:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171029220234AAkW6Nl