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March 5th, 2019, 12:20 AM   #1
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How can I solve Logarithmic equations with multiple depented terms?

I know how to solve this:

$\displaystyle
log_2(x) + log_2(x+2) = 2
$

But how do I solve this:
$\displaystyle
8n^2 - 64nlog_2(n) = 0
$

If I try to get rid of the log:
$\displaystyle
64nlog_2(n) = 8n^2 \Leftrightarrow 2^{64nlog_2(n)} = 2^{8n^2} \Leftrightarrow
2^{log_2(n^{64n})} = 2^{8n^2} \Leftrightarrow n^{64n} = 2^{8n^2}
$
Dead end...

I have years to deal with these kind of equations and I don't remember what are the tools and the rules to solve them. Can someone remind me?

Thank you.
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March 5th, 2019, 07:34 AM   #2
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$8n^2 - 64n \log_2(n) = 0$

$8n^2 = 64 n \log_2(n)$

$\text{as }n \text{ is the argument of the }\log_2()\text{ function }n \neq 0$

$8n = 64 \log_2(n)$

$\dfrac{n}{8} = \log_2(n)$

$2^{n/8} = n$

This doesn't have a nice closed form solution in the usual vocab of functions.

It's solution can be expressed using the Lambert W function.

Numerically

$x \approx 1.099997,~43.55926$
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March 5th, 2019, 11:02 AM   #3
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Quote:
Originally Posted by romsek View Post
$8n^2 - 64n \log_2(n) = 0$

$8n^2 = 64 n \log_2(n)$

$\text{as }n \text{ is the argument of the }\log_2()\text{ function }n \neq 0$

$8n = 64 \log_2(n)$

$\dfrac{n}{8} = \log_2(n)$

$2^{n/8} = n$

This doesn't have a nice closed form solution in the usual vocab of functions.

It's solution can be expressed using the Lambert W function.

Numerically

$x \approx 1.099997,~43.55926$
So you actually dividing with n both sides of the equation with the assumption that n is not zero?
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March 5th, 2019, 12:53 PM   #4
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Quote:
Originally Posted by babaliaris View Post
So you actually dividing with n both sides of the equation with the assumption that n is not zero?
If $\log_2(n)$ is valid then $n\neq 0$
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March 5th, 2019, 12:55 PM   #5
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n cannot be zero .
Before starting to solve an equation the domain must always be found first .
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