What does a $\ln$ transformation do to an inequality?

Question

I am trying to learn more about what the $\ln$ transformation does to a simple inequality like $a>b$. Is it true that $\ln(a)>\ln(b)$?

Background: I am trying to solve

$$\exp(-\frac{1}{2}(1-x)^2)-\exp(-\frac{1}{2}5 x^2) >0$$

for $x$, where (importantly for the following) $x$ is $0$ or $1$.

Edit: After comments I correct the inequalities to

$$ \exp(-\frac{1}{2}(1-x)^2)-\exp(-\frac{1}{2}x) >0 \\ \exp(-\frac{1}{2}(1-x)^2) > \exp(-\frac{1}{2}x) \\ (1-x)^2 < x \\ 1-2x+x < x \\ x > \frac{1}{2}.$$

and thus $x=1$ given its parameter space.

Answer 1

The logarithm is a strictly increasing function, and therefore applying it to both sides of an inequality preserves the inequality.

Just be sure you know that both sides of the inequality are positive, such that the logarithm is defined at all.

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For your concrete peoblem, it would probably be easier to start by rearraning to $$ e^{-\frac12(1-x)^2} > e^{-\frac12x^2} $$ then immediately take logarithms, and be left with the plain old algebraic inequality $$ -\tfrac12(1-x)^2 > -\tfrac12x^2 $$

Answer 2

$\ln x$ is a strictly increasing function, so $\ln(a)>\ln(b)\iff a>b$.

However, you also say that your original expression can be simplified to $\exp(1-2x)>1$. I don't see how this expression can be equivalent to the original.