Expand $\ln(y)$ where $ y=x^3-5x^2+3x+9 $ and find its derivative

Question

I have to expand the polynomial $$ y=x^3-5x^2+3x+9 $$ with a natural logarithm and then take the derivative.

In my opinion $\ln(y)$ is $$ \ln(y) =3\ln(x)-10\ln(x)+\ln(3x)+\ln(9). $$ So the derivative is $$ (\ln(y))' =\frac{3}{x}-\frac{10}{x}+\frac{3}{3x}. $$

It's correct?

Answer 1

I don't agree with you. $$\ln{y}=\ln(x^3-5x^2+3x+9)$$ and $$\left(\ln{y}\right)'=\frac{3x^2-10x+3}{x^3-5x^2+3x+9}.$$

Answer 2

This is not true, we have that $$\log(ab) =\log(a)+\log(b) \qquad \text{and} \qquad \log\left(\frac{a}{b}\right) =\log(a)-\log(b). $$

Answer 3

Remember the laws of logs

$$y=x^3-5x^2+3x+9$$

$$\ln(y)=\ln(x^3-5x^2+3x+9)$$

This cannot be simplified because $\ln(A+B)\neq \ln(A)+\ln(B)$

Note that $\ln(AB)=\ln(A)+\ln(B)$.

Answer 4

You made a mistake assuming that in addition of its special property $\ln(ab)=\ln(a)+\ln(b)$, the logarithm was also linear.

But there could not be such function. Indeed if we suppose there exists a function $f$ verifying $$\begin{cases}f(x+y)=f(x)+f(y)\\f(xy)=f(x)+f(y)\end{cases}$$

Then let's call $a=f(1)$.

• $f(n)=f(1)+\cdots+f(1)=na$
• $a=f(\frac nn)=f(n)+f(\frac 1n)\iff f(\frac 1n)=(1-n)a$
• for rationals $f(\frac pq)=pa+(1-q)a=(1+p-q)a$

But this is where it gets problematic, let's have $p>0,q>0,p\neq q$:

$f(\frac pq)=f(\frac{2p}{2q})\implies (1+p-q)a=(1+2p-2q)a\implies (-p+q)a=0\implies a=0$

So $f(x)=0$ on positive rationals which is not a very interesting function after all.

Note: I dismissed voluntarily exploiting $f(0)$ because $\ln$ not defined in $0$, but you get a shortcut if you use that ($f(0)=2f(0)$ then $f(0)=0$ then $f(0)=f(x)+f(0)$ so $f(x)=0$ for all $x$).

Answer 5

If that was the derivative of $\ln y$, then you could simplify it to $$ (\ln y)'=-\frac{6}{x} $$ which would mean $\ln y=k-6\ln x$. Evaluating at $x=1$, $\ln 8=k$. Do you believe this? I don't think so.

The exercise rather asks you to realize that $$ x^3-5x^2+3x+9=(x+1)(x-3)^2 $$ so the expression is positive for $x>-1$, but $x\ne3$. Over the domain $(-1,3)\cup(3,\infty)$ the logarithm can be computed and $$ \ln y=\ln\bigl((x+1)(x-3)^2\bigr)=\ln(x+1)+2\ln\lvert x-3\rvert $$ and therefore $$ (\ln y)'=\frac{1}{x+1}+\frac{2}{x-3} $$