Let's say I want to leave $x$ alone in this inequality: $$-2 \ln x \leq y$$ where $x,y \in \Bbb R$.
Well one way of doing it is first dividing the $-2$ by both sides --> $$\ln x \geq -\frac{1}{2}y$$ and then taking $e$ and raising it to the power of the left and right hand sides --> $$x \geq e^{-\frac{1}{2}y}$$
But another way of leaving $x$ alone is by first using the power rule of the natural log function: $a\ln b = \ln (b^{a})$, thus we have --> $$\ln (x^{-2}) \leq y$$ and then taking $e$ and raising it to the power of the left and right hand sides --> $$x^{-2} \leq e^{y}$$ and finally raising each side to the exponent $-\frac{1}{2}$ --> $$x \leq e^{-\frac{1}{2}y}$$
I guess my question is what did I do wrong haha.
For strictly-increasing $h,$ $$f(x)<g(x)\implies h\big(f(x))<h(g(x)\big).$$ For strictly-decreasing $h,$ $$f(x)<g(x)\implies h\big(f(x))>h(g(x)\big).$$