NO CALCULATOR: solve for x. Been doing this for about half an hour and I've been going around in circles.?
回答 (12)
2019-11-03 7:52 am
just use a logarithm and then figure it out....... so easy
2019-11-05 12:50 am
= 3 log(2) / [ 3log(2)-log(5) ]
2019-11-03 1:09 pm
(1/2)^(3/x) = 5/8 = 5*(1/2)^3
(3/x)*[log2(1/2)] = log2(5) + 3*log2(1/2)
(3/x)(-1) = log2(5) + 3*(-1)
-(3/x) = log2(5) - 3
3/x = 3 - log2(5)
x = 3/(3 - log2(5)) … {that's log-to-the-base-two}
(3/x)*[log2(1/2)] = log2(5) + 3*log2(1/2)
(3/x)(-1) = log2(5) + 3*(-1)
-(3/x) = log2(5) - 3
3/x = 3 - log2(5)
x = 3/(3 - log2(5)) … {that's log-to-the-base-two}
2019-11-03 7:57 am
Did you try taking the logarithm of each side? x = log(8) / log( 8/5). Without a calculator, that's about x = 4.4...
2019-11-03 7:55 am
(3/x) * ln(1/2) = ln(5/8)
x * ln(5/8) = 3 * ln(1/2)
x = 3 * ln(1/2) / ln(5/8)
x = ln(1/8) / ln(5/8)
x = -ln(8) / (ln(5) - ln(8))
x = ln(8) / (ln(8) - ln(5))
x = ln(8) / ln(8/5)
x = ln(8) / ln(1.6)
x = 3 * ln(2) / ln(1.6)
That's about as nice as it will get. "No Calculator" doesn't necessarily mean, "give us a decimal expression." It could just mean, "don't use a calculator to evaluate this problem."
x * ln(5/8) = 3 * ln(1/2)
x = 3 * ln(1/2) / ln(5/8)
x = ln(1/8) / ln(5/8)
x = -ln(8) / (ln(5) - ln(8))
x = ln(8) / (ln(8) - ln(5))
x = ln(8) / ln(8/5)
x = ln(8) / ln(1.6)
x = 3 * ln(2) / ln(1.6)
That's about as nice as it will get. "No Calculator" doesn't necessarily mean, "give us a decimal expression." It could just mean, "don't use a calculator to evaluate this problem."
2019-11-03 4:41 pm
(1/2)^(3/x) = 5/8
(1/2)^3 = (5/8)^x
(5/8)^x = 1/8
x = (ln(1/8) + 2*i*pi*n) / ln(5/8), for any integer n
x =~ 4.4243095420708459538464400889368 + 13.36837614905841750962828210159*i*n, for any integer n
(1/2)^3 = (5/8)^x
(5/8)^x = 1/8
x = (ln(1/8) + 2*i*pi*n) / ln(5/8), for any integer n
x =~ 4.4243095420708459538464400889368 + 13.36837614905841750962828210159*i*n, for any integer n
2019-11-05 11:56 pm
It says no calculator, but doesn't say no paper and pencil.
My first instinct on seeing the 2 and 8 is to take the log base 2 (log2) of both sides.
-1 * 3/x = log2(5/8) = log2(5) - log2(8) = log2(5) - 3
x = 3/(3 - log2(5))
That's the exact answer. If they had asked for a numerical approximation, log2(5) we know is slightly more than 2, since 2^2 = 4, but a lot less than 3, since 2^3 = 8. If we have memorized that sqrt(2) = 1.4 approx, then we know that 4 * 1.4 = 5.6, so the log2(5) must be less than 2.5. Or even better, if we memorized that the cube root of 2 is about 1.26, then since 4*1.25 = 5, log2(5) must be darn close to 2 + .33 = 2.33. If the square or cube roots of 2 were not memorized, then one would have to take the roots by hand, but that skill is even rarer than memorizing the values.
My first instinct on seeing the 2 and 8 is to take the log base 2 (log2) of both sides.
-1 * 3/x = log2(5/8) = log2(5) - log2(8) = log2(5) - 3
x = 3/(3 - log2(5))
That's the exact answer. If they had asked for a numerical approximation, log2(5) we know is slightly more than 2, since 2^2 = 4, but a lot less than 3, since 2^3 = 8. If we have memorized that sqrt(2) = 1.4 approx, then we know that 4 * 1.4 = 5.6, so the log2(5) must be less than 2.5. Or even better, if we memorized that the cube root of 2 is about 1.26, then since 4*1.25 = 5, log2(5) must be darn close to 2 + .33 = 2.33. If the square or cube roots of 2 were not memorized, then one would have to take the roots by hand, but that skill is even rarer than memorizing the values.
2019-11-04 1:38 pm
(1/2)^(3/x) = 5/8
(1/8)^(1/x) = 5/8
1/8 = (5/8)^x
0.125 = 0.625^x
(1/8)^(1/x) = 5/8
1/8 = (5/8)^x
0.125 = 0.625^x
2019-11-03 11:58 am
(1/2)^(3/x) = 5/8. Raise both side to the power x.
(1/2)^3=(5/8)^x= 1/8.
You have now 5^x= 8^(x-1)
(1/2)^3=(5/8)^x= 1/8.
You have now 5^x= 8^(x-1)
2019-11-03 8:38 am
(1/2)^(3/x) = 5/8
[(1/2)^3]^(1/x) =5(1/2)^3
[(1/2)^3]^(1/x – 1)) = 5
(1/x – 1)log(1/8) = log5
1/x = -(log5)/log8 + 1
x = log8/(log8 – log5)
= 3log2/(3log2 + log2 -1)
= 3log2/(4log2 -1)
ALL THE ABOVE LOGS ARE TO BASE 10.
[(1/2)^3]^(1/x) =5(1/2)^3
[(1/2)^3]^(1/x – 1)) = 5
(1/x – 1)log(1/8) = log5
1/x = -(log5)/log8 + 1
x = log8/(log8 – log5)
= 3log2/(3log2 + log2 -1)
= 3log2/(4log2 -1)
ALL THE ABOVE LOGS ARE TO BASE 10.
收錄日期: 2021-04-11 23:01:49
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