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

AI Thread Summary
The discussion centers on proving the equation log(x)·log(x^2)·log(x^3)·...·log(x^90) = 4095. A participant points out that the equation is false when x=1, leading to a contradiction. The correct interpretation involves summing the logarithms, which can be expressed as log(x) multiplied by the sum of integers from 1 to 90. This results in the equation summing to 4095·log(x), clarifying the misunderstanding. The final conclusion emphasizes the importance of correctly interpreting logarithmic identities.
Chipset3600
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Hello MHB.

How can i proof this equation?

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

Chipset3600 said:
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.
 
Re: Proof

Chipset3600 said:
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.
 
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