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If $\log 0.318 = x$ and $\log 0.317 = y$, can $\log 0.319$ be expressed in terms of $x$ and $y$ ?

Is there any way or we have to find $\log 0.319$ using log tables only?

I'm not getting any expression in $x, y$ which will represent $\log 0.319$.

Please help.

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Approximately or exactly? (If the former, then yes; if the latter, then no, at least not in a way which doesn't require the introduction of some other irrational number.) – Qiaochu Yuan Jul 17 '12 at 5:46
Approximately. Taking it further, $x=0.3364$ and $y=0.3332$. – TheApe Jul 17 '12 at 5:48
Think about linearly approximating the logarithm. – Qiaochu Yuan Jul 17 '12 at 5:51
We can do linear extrapolation. The $\log$ tables that I used as a student gave $\log w$ for $w$ listed to only $4$ places. They also had tables of "proportional parts" to make interpolation or extrapolation easier. – André Nicolas Jul 17 '12 at 5:56
up vote 6 down vote accepted

The question clearly hints at the following reasoning.

One is given $x=u(t+h)$ and $y=u(t)$, where $t=0.317$, $h=0.001$, and $u$ is the logarithm function, and one is asked an approximation of $z=u(t+2h)$.

Since $h$ is small, $x\approx y+hs$ where $s=u'(t)$ is the slope of the function $u$ at point $t$ (whose value will be irrelevant). Likewise, $z\approx y+2hs$. Solving yields $z\approx 2x-y$.

The error is of order $h^2=0.000001$, to be compared to the suggested value $2x-y\approx x$, which is of order $1$.

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Did, your answer here :….... is the BEST and WITTIEST answer ever given in the history of maths! – Bazinga Jul 17 '12 at 12:07
@Bazinga: This appreciation is clearly overrated. But thanks anyway... – Did Jul 17 '12 at 12:40

One has $0.318=e^x$, $0.317=e^y$ and $0.319=2\times0.318-0.317$. Thus, $0.319 = 2e^x-e^y$ and

$$\log 0.319 = \log( 2e^x-e^y)$$

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I have put your answer into TeX. Mostly, this just amounted to enclosing all the math in dollar signs. You can see exactly what I did by clicking on "edit". If you're going to spend much time here, it would be good if you could learn the rudiments of TeX. – Gerry Myerson Jul 17 '12 at 6:02
Here $\log 0.319$ is indeed expressed in terms of $x$ and $y$, but not in a way that enables us to compute $\log 0.319$ easily. – André Nicolas Jul 17 '12 at 6:05
This obviously DOES NOT answer the question (and illustrates once again what can happen when accepting answers barely 20 minutes after the question was asked...). – Did Jul 17 '12 at 11:29

Using second-order central differences, we have $$ f''(t) \approx \frac{f(t+h) - 2 f(t) + f(t-h)}{h^{2}} $$

Let $f=\log$, $t=0.318$, $h=0.001$, and $z=\log 0.319$. Then $$ -\frac{1}{0.318^2}\approx \frac{z-2x+y}{0.001^{2}} $$ Solving for $z$ we get $$ z \approx 2x-y -\frac{0.001^{2}}{0.318^2} $$ This is actually a very good approximation for $z$.

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See if this also helps :

Let $x_1=0.318$ $y_1=0.317$ $z_1=0.319$ Hence we have $\sqrt{x_1z_1}=0.31799842\approx y_1 $ $$\Rightarrow \frac{1}{2}(logx_1+logz_1)\approx logy_1 $$ $$\Rightarrow logz_1\approx2logy_1-logx_1$$ $$\Rightarrow log 0.319=2y-x $$

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