(1/2)^(3/x) = 5/8
Is there some intuitive algebraic manipulation that anyone here can find that would give you the answer without logs? Here's the direction I take:
(1/2)^(3/x) = 5/8
2^(-3/x) = 5 * 2^(-3)
2^(-3) = 5^x * 2^(-3x)
2^(-3+3x) = 5^x
8^(-1) * 8^x = 5^x
8^(x-1) = 5^x
x = 4.424
My point is that I have to use a calculator and find the answer. I can use logs but I don't know ln of 5/8 off the top of my head. Everyone in my last question did something similar to this, but I want to know if it's possible to manipulate this in a more optimal way so that you don't need a calculator and can still get the answer 4.424. Thanks
NO CALCULATOR: can you find an exact value for x without a calculator and give a final answer in numerical form (ie no 'ln's allowed)?
2019-11-03 10:53 pm
回答 (3)
2019-11-03 11:47 pm
✔ 最佳答案
I read that Newton knew a method for finding logs by hand and use to do it 'for fun'. None of my math classes ever taught how, so I don't know.=== All your work is great -- as you've probably guessed, I'm old. Do you consider a slide rule as a calculator? You can use those for logs. Otherwise, I see no way around using a calculator for the logs.
2019-11-04 1:48 pm
(1/2)^(3/x) = 5/8
(1/2)^(3/x) = 5/8
2^(-3/x) = 5 * 2^(-3)
2^(-3) = 5^x * 2^(-3x)
2^(-3+3x) = 5^x
8^(-1) * 8^x = 5^x
8^(x-1) = 5^x
x = 4.424
(1/2)^(3/x) = 5/8
2^(-3/x) = 5 * 2^(-3)
2^(-3) = 5^x * 2^(-3x)
2^(-3+3x) = 5^x
8^(-1) * 8^x = 5^x
8^(x-1) = 5^x
x = 4.424
2019-11-04 12:05 am
Tahing things from your 8^(x-1) = 5^x:
8^x/8= 5^x
(8/5)^x = 8, 1.6^x = 8
From her on it's all numerical techniques.
It is easy to establish that x > 4 and x < 5.
So maybe you can try to home in on x by
repeated bisection of an appropriate
sequence of sub-interval of (4, 5).
8^x/8= 5^x
(8/5)^x = 8, 1.6^x = 8
From her on it's all numerical techniques.
It is easy to establish that x > 4 and x < 5.
So maybe you can try to home in on x by
repeated bisection of an appropriate
sequence of sub-interval of (4, 5).
收錄日期: 2021-04-11 22:59:41
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