How Do You Prove a Logarithmic Identity Involving Powers of x?

[Chipset3600]

Hello MHB.

How can i proof this equation?

[h=5]log(x).log(x^2).log(x^3)... log(x^90)=4095[/h]

 

[caffeinemachine]

Re: Proof

Hello MHB.

How can i proof this equation?

log(x).log(x^2).log(x^3)... log(x^90)=4095

This ain't true. Put x=1 and you get 0=4095.

 

[TheBigBadBen]

Re: Proof

Hello MHB.

How can i proof this equation?

[h=5]log(x).log(x^2).log(x^3)... log(x^90)=4095[/h]

Let's assume you mean to do something with the expression
$$log(x)+log(x^2)+log(x^3)+\cdots +log(x^{90})$$
(by the way, we could also write this as $\sum_{n=1}^{90} log\left(x^n\right)$ )

What we could do is say that for any n, we have $log(x^n)=n\cdot log(x)$. With that in mind, our sum becomes
$$log(x)+2 \, log(x)+3\, log(x)+\cdots +90\, log(x)$$
factoring, we have
$$(1+2+3+\cdots+90)\,log(x)$$
which gives us
$$\sum_{n=1}^{90} log\left(x^n\right)=\frac{91\cdot 90}{2} \,log(x) = 4095\,log(x)$$
Which is what I assume you meant.