Natural Logarithm can't understand properties

Question

~I don't get some of the properties of natural logarithm ($\ln$).

$\ln(x^y) = y\ln(x)$

ex. $3\ln 7 = \ln 343$

and what is the difference between the above example and this

$3\ln^2(7)$ not equal to $(\ln(7^3))^2$ and how can you simplify it?

$\ln(x) - (\ln(y)) = \ln(x/y)$

~Why $\ln(9/2)$ not equal to $\ln(9)/(\ln(2))$

and why $\ln(\ln(8/3))$ not equal to $\ln(\ln(8))- (\ln(\ln(3))$

$\ln(\ln(8))- (\ln(\ln(3)) = \ln(\ln(8)/(\ln(3))$

Please help me.

Answer 1

Let't try this one: $3\ln^2(7)$ not equal to $(\ln(7^3))^2$

In fact, $$ (\ln(7^3))^2 = (3\ln 7)^2 = 3^2 (\ln 7)^2 = 9 \ln^2 7 $$

Answer 2

In general, for a function $$f(x+y)\ne f(x)+f(y),$$ $$f(x-y)\ne f(x)-f(y),$$ $$f(x\cdot y)\ne f(x)\cdot f(y),$$ $$f(\frac xy)\ne \frac{f(x)}{f(y)}.$$

The logarithm enjoys the very special property that

$$f(x\cdot y)=f(x)+f(y),$$

which makes it extremely useful as it turns the multiplies into additions. There are related identities $$f(\frac xy)=f(x)-f(y),$$and $$f(x^y)=y\cdot f(x).$$ There is no formula about $f(x+y)$ nor $f(x-y)$, that's it.

The fundamental property can be understood by remembering that the logarithm is the inverse of the exponential, $g(x)=e^x$.

You known that the exponential turns a sum into a product$$g(X+Y)=e^{X+Y}=e^X\cdot e^Y=g(X)\cdot g(Y).$$

The logarithm does the inverse, as shown by $$g(f(x)+f(y))=g(f(x))\cdot g(f(y))$$ $$f(g(f(x)+f(y)))=f(x)+f(y)=f(g(f(x))\cdot g(f(y)))=f(x\cdot y).$$

Answer 3

Old question, I know, but I don't feel there is a complete answer yet.

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$\ln(x^y) = y\ln(x)$

Let $M = x^y$. Then $y = \log_x M$ by the definition of logarithm. Consider then: $$y \ln x = (\log_x M) \cdot (\ln x) = \frac{\ln M}{\ln x} \cdot \ln x$$ This last equality above comes from the change of base formula. Simplifying the result gives us $y\ln x = \ln M$. And since $M = x^y$, then we have $y\ln x = \ln(x^y)$.

and what is the difference between the above example and this

$3\ln^2(7)$ not equal to $(\ln(7^3))^2$ and how can you simplify it?

The difference is that you now have $\ln^2$ and not just $\ln$. It's unclear to me if by $\ln^2(7)$ you mean $\ln (\ln 7)$ or $\left(\ln 7\right)^2$, so here it is both ways, each using $y\ln x = \ln(x^y)$.

$$ 3\ln^2(7) = 3\ln(\ln 7) = \ln([\ln 7]^3)$$ For that one, we used $y\ln x = \ln(x^y)$ with $y=3$ and $x = \ln 7$.

$$ 3\ln^2(7) = 3[\ln 7]^2 = \sqrt{3}^2 [\ln 7]^2 = [\sqrt3 \ln 7]^2 = \left[\ln(7^{\sqrt3})\right]^2$$ For that one, we used $y\ln x = \ln(x^y)$ with $y=\sqrt3$ and $x = 7$.

Either way, I would argue that $3\ln^2 7$ is actually the more simplified version. But it's important to clarify what you mean by $\ln^2(7)$ and perhaps either use $\ln(\ln 7)$ or $(\ln 7)^2$.

~Why $\ln(9/2)$ not equal to $\ln(9)/(\ln(2))$

Because it isn't. My counter-question would be why would you think they'd be equal? You even put the correct formula in the line above it:

$\ln(x) - (\ln(y)) = \ln(x/y)$

For a more concrete reason why $\ln(9/2) \ne (\ln9)/(\ln2)$, note that $\ln(9/2) \approx 1.5040$ and $(\ln9)/(\ln2) \approx 3.1699$. Were you perhaps asking why $\ln x - \ln y = \ln(x/y)$ in general?

and why $\ln(\ln(8/3))$ not equal to $\ln(\ln(8))- (\ln(\ln(3))$

Same answer as before: Because they aren't. And why would you think they would be? $\ln(\ln(8/3)) \approx -0.0194$ and $\ln(\ln8) - \ln(\ln3) \approx 0.6381$.