$x \sim y \implies \log x \sim \log y$? Does $x \sim y \implies \ln x \sim \ln y$?
I would have thought not, but the following has almost persuaded me otherwise: 
Assume $x \sim y.$ Does this imply that $$\tag{1}I = \frac{\int_{1}^{x}\frac{dt}{t}}{\int_{1}^{y}\frac{dt}{t}}\sim 1?$$
WLOG let $(y - x) = d, d > 0.$ At worst the difference of the integrals will be less than $(1/x)(y-x).$  
Then $$ I =  \frac{\ln x}{\ln x + (y-x)(\frac{1}{x})} = \frac{\ln x }{\ln x + \frac{y}{x} -1}. $$
Since $\frac{y}{x}\sim 1,$ $$ \lim_{x \to \infty} I = 1.$$     
Is this correct (and if so is there a better proof)?  Thanks.
 A: That follows only if $y\to \infty$ (or, more precisely, if $\log(y)$ does not tend to zero). Under this assumption (that you implicitely used in your last formula), we get:
$$\frac{x}{y} \to 1 \implies \log\left(\frac{x}{y}\right)=\log(x)-\log(y)\to 0  \implies \frac{\log(x)}{\log(y)}\to 1$$
BTW, the last implication is not reversible, and the converse is not true.
A: I took the question to mean:  if $x(t)/y(t) \to 1$ as $t \to \infty$, does it follow that $\log(x(t))/\log(y(t))\to 1$ also?  Then, as leonbloy hints, we can get a counterexample,
$$
x(t) = 1+\frac{1}{t},\qquad y(t) = 1+\frac{1}{t^2} .
$$
As the other answers show, there is no such counterexample if also $x(t) \to \infty$.
A: By the definition, $y/x\to 1$ as $x\to\infty$, you get that for any $\epsilon>0$ there is an $N$ such that if $x>N$ $y/x\in (1-\epsilon,1+\epsilon)$, or $y\in ((1-\epsilon)x,(1+\epsilon)x)$.
Taking the log of both sides gives you:
$$\log (1-\epsilon) + \log x < \log y < \log(1+\epsilon) +\log x$$
Dividing by $\log x$, you get, for $x>N$:
$$1+\frac{\log (1-\epsilon)}{\log x} < \frac {\log y}{\log x} < 1 + \frac{\log(1+\epsilon)}{\log x}$$
Now show that this implies your condition.
