If $\log 0.318 = x$ and $\log 0.317 = y$, can $\log 0.319$ be expressed in terms of $x$ and $y$? 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.
 A: 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 $1$.
A: 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 $z$.
A: 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
$$\log 0.319 = \log( 2e^x-e^y)$$
A: 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 $ $$\Rightarrow \frac{1}{2}(logx_1+logz_1)\approx logy_1 $$ $$\Rightarrow logz_1\approx2logy_1-logx_1$$ $$\Rightarrow log 0.319=2y-x $$
