Logarithms 10pts best explanation?
2016-03-30 2:06 am
10^-x = 5^2x we have to do change of base formula, but what do we change the base to & why ?
回答 (3)
2016-03-30 2:14 am
10^(-x) = 5^(2x)
log[10^(-x)] = log[5^(2x)]
-x log(10) = 2x log(5)
-x = 2x log(5)
2x log(5) + x = 0
x [2 log(5) + 1] = 0
x = 0
log[10^(-x)] = log[5^(2x)]
-x log(10) = 2x log(5)
-x = 2x log(5)
2x log(5) + x = 0
x [2 log(5) + 1] = 0
x = 0
2016-03-30 2:10 am
10^(-x) = 5^(2x)
Let's get the log of both sides:
ln[10^(-x)] = ln[5^(2x)]
Now move the exponents out of the logs:
-x ln(10) = 2x ln(5)
Move both terms to the same side so we can factor out the x:
0 = 2x ln(5) + x ln(10)
0 = x[2 ln(5) + ln(10)]
Divide both sides by the constant (2 ln(5) + ln(10)):
x = 0
Let's get the log of both sides:
ln[10^(-x)] = ln[5^(2x)]
Now move the exponents out of the logs:
-x ln(10) = 2x ln(5)
Move both terms to the same side so we can factor out the x:
0 = 2x ln(5) + x ln(10)
0 = x[2 ln(5) + ln(10)]
Divide both sides by the constant (2 ln(5) + ln(10)):
x = 0
2016-03-30 2:09 am
5^(2x)=10^-x
25^x = 1/10^x
(10^x)(25^x) = 1
(10*25)^x=1
250^x = 1
---> x=0
25^x = 1/10^x
(10^x)(25^x) = 1
(10*25)^x=1
250^x = 1
---> x=0
收錄日期: 2021-04-18 14:40:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160329180620AARd226