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Help with equation....thank you?

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  • Help with equation....thank you?


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Answer #1 | 30/12 2013 06:15
First lets make z = 5^(x-5) Then we can change this to 2z - 6z/5 = 25 multiply both sides by 5 and get 10z - 6z = 125 4z = 125 z = 125/4 So 5^(x-5) = 125/4 take log base 5 of both sides and get (x - 5) = log(5) (125/4) x = log(5) (125/4) + 5 = 7.138647
Positive: 100 %
Answer #2 | 30/12 2013 06:09
2 * 5^(x - 5) - 6 * 5^(x - 6) = 5^2 2 * 5^(x) / 5^(5) - 6 * 5^(x) / 5^(6) = 5^2 2 * 5^(x) * 5 - 6 * 5^(x) = 5^(2) * 5^(6) 10 * 5^(x) - 6 * 5^(x) = 5^8 4 * 5^(x) = 5^(8) 5^(x) = (1/4) * 5^8 x * ln(5) = ln(1/4) + 8 * ln(5) x * ln(5) = 8 * ln(5) - ln(4) x = (8 * ln(5) - ln(4)) / ln(5)
Positive: 0 %
Answer #3 | 30/12 2013 06:13
2 * 5^(x - 5) - 6 * 5^(x - 6) = 5^2 2 * 5^x/5^5 - 6 * 5^x/5^6 = 5^2 2 * 5 * 5^x - 6 * 5^x = 5^8 4 * 5^x = 5^8 5^x = 5^8 / 4 x = log₅(5^8/4) x = 8 - log₅(4)
Answer #4 | 30/12 2013 06:49
2.5^(x-5) - 6.5^(x-6) = 5^2 (5^(x-6))(10 - 6) = 5^2 5^(x-8) = 1/4 x - 8 = ln(1/4) / ln(5) x = 8 - ln(4)/ln(5) = 7.138646883853214
Positive: 0 %

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