how do I solve the following problem:
given log(x) for x=310,320,330,340,350 and 360 as 2.4913,2.5051,5.5185,2.5440 and 2.5563 find the value of log 337.5
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sid597
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Possible duplicate of An alternative way to calculate log(x) – Jul 14 '16 at 19:16
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@KingDuken can you explain how is it so? – sid597 Jul 14 '16 at 19:18
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1Linear approximation claims $f (x) - f(a) \approx f'(a)(x-a) $ so $340 - \log (337.5) \approx \frac {1}{337.5\ln 10}(340 - 337.5) $ – fleablood Jul 14 '16 at 19:18
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Can't you plot a graph of the values you have and then interpolate the value for 337.5 ? This would use all the values you have been given, so you should get a 'good feel' for the curve of the graph in that region so you'll have a better approximation. At what level is the answer expected? – user247608 Jul 14 '16 at 20:23
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how do you calculate ln10? – user247608 Jul 14 '16 at 20:29
2 Answers
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Just playing around with the numbers, we get that the geometric mean of 310, 330, 340, 350, 360 (and leaving out 320) is about 337.55536830584776. Thus, $\log 337.5$ is roughly the arithmetic mean of their given logs, or 2.5283.
Of course, we can do a lot better with various products of the $x$ values (and the associated linear combination of the logs), but that's close enough to be correct to 4 significant figures (the true value is about 2.5282737771670436).
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+1. I assume by "various products" you meant "various powers"; I followed up on that suggestion in my own answer. – Brian Tung Jul 14 '16 at 20:28
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Following up on Mnemonic's discussion, just two of the five provided log values are needed. We have
$$ 320 = 2^6 \times 5 $$
and
$$ 360 = 2^3 \times 3^2 \times 5 $$
and
$$ 337.5 = 2^{-1} \times 3^3 \times 5^2 $$
and of course
$$ 10 = 2 \times 5 $$
This gives us
$$ 337.5 = \frac{360^{3/2}\times 10^{17/10}}{320^{6/5}} $$
Since $\log 10 = 1$, we get
$$ \log 337.5 = \frac{3}{2} \log 360 + \frac{17}{10} - \frac{6}{5} \log 320 $$
which is as accurate as the log values we have. With the log values as given, we obtain
$$ \log 337.5 \doteq 1.5 \times 2.5563 + 1.7 - 1.2 \times 2.5051 \doteq 2.5283 $$
the same answer as obtained by Mnemonic. However, although $\log 360 \doteq 2.55630250$, so the provided value is fairly accurate, $\log 320 \doteq 2.50514998$, which does indeed round down to $2.5051$, but only barely. Using just one more significant digit yields $2.52827$.
Brian Tung
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