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- #1

#### Chipset3600

##### Member

- Feb 14, 2012

- 79

Hello MHB.

How can i proof this equation?

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

How can i proof this equation?

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

- Thread starter Chipset3600
- Start date

- Thread starter
- #1

- Feb 14, 2012

- 79

Hello MHB.

How can i proof this equation?

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

How can i proof this equation?

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

- Mar 10, 2012

- 835

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

How can i proof this equation?

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

- May 12, 2013

- 84

Let's assume you mean to do something with the expressionHello MHB.

How can i proof this equation?

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

$$log(x)+log(x^2)+log(x^3)+\cdots +log(x^{90})$$

(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.

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