Let $p(n)$ the predicate $\ln{x}^{n}=n\ln{x}$. Show the implication $p(n)\Rightarrow p(n+1)$ for all natural number $n$. I need your help with this exercise. I don't know if what I did is correct.
Let $p(n)$ the predicate $\ln{x}^{n}=n\ln{x}$. Show the implication $p(n)\Rightarrow p(n+1)$ for all natural number $n$.
Shall I use math induction?
If $n=1$, then
$$ln{x}^{1}=1\ln{x}$$
$$ln{x}=\ln{x}$$
Therefore the statement is true for $n=1$.
If $n=k$ is true, then para $n=k+1$ is true.
Let's suppose that
$$\ln{x}^{k}=k\ln{x}$$
Then
$$\ln{x}^{k+1}=(k+1)\ln{x}$$
$$\ln({x^{k}\cdot x^{1}})=(k+1)\ln{x}$$
$$\ln{x^{k}}+\ln{x^{1}}=(k+1)\ln{x}$$
$$\ln{x^{k}}+\ln{x}=k\ln{x}+\ln{x}$$
$$k\ln{x}+\ln{x}=k\ln{x}+\ln{x}$$
Therefore, the statement is true for $n=k+1$.
Therefore, the statement is true for all natural numbers.
 A: There are two things here :
[[1]] Show that $P(n)$ is true for all $n$.
[[ this is not Explicit in the Question which is not asking to Prove this ]]
[[2]] Show that $P(n) \implies P(n+1)$.
[[ this is Explicit in the Question which is asking to Prove Exactly this ]]
We can show [[1]] in various ways, including Induction where we have to show $P(n)$ for $(n=1)$ [[ that is : we have to show $P(1)$ ]] & then show that $P(n) \implies P(n+1)$.
You have done this but in reverse :
Start with $k\ln{x}+\ln{x}=k\ln{x}+\ln{x}$
then move to $\ln{x}^{k+1}=(k+1)\ln{x}$
Here it is :
We have $n=k$ here :
$k\ln{x}=\ln{x^{k}}$ [[ Equation 1 ]]
Now, we start with :
$k\ln{x}+\ln{x}=k\ln{x}+\ln{x}$
$\ln{x^{k}}+\ln{x}=k\ln{x}+\ln{x}$ [[ using Equation 1 ]]
$\ln{x^{k}}+\ln{x^{1}}=(k+1)\ln{x}$
$\ln({x^{k}\cdot x^{1}})=(k+1)\ln{x}$
$\ln{x}^{k+1}=(k+1)\ln{x}$
We have got $n=k+1$ here :
$(k+1)\ln{x}=\ln{x^{k+1}}$ [[ Equation 2 ]]
We have shown that Equation 1 Implies Equation 2
That is DONE !
We can show [[2]] with only that Part & not worry about $P(1)$.
