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 March 5th, 2019, 12:20 AM #1 Senior Member   Joined: Oct 2015 From: Greece Posts: 107 Thanks: 6 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.
 March 5th, 2019, 07:34 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,373 Thanks: 1276 $8n^2 - 64n \log_2(n) = 0$ $8n^2 = 64 n \log_2(n)$ $\text{as }n \text{ is the argument of the }\log_2()\text{ function }n \neq 0$ $8n = 64 \log_2(n)$ $\dfrac{n}{8} = \log_2(n)$ $2^{n/8} = n$ 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 $x \approx 1.099997,~43.55926$ Thanks from topsquark and idontknow
March 5th, 2019, 11:02 AM   #3
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Quote:
 Originally Posted by romsek $8n^2 - 64n \log_2(n) = 0$ $8n^2 = 64 n \log_2(n)$ $\text{as }n \text{ is the argument of the }\log_2()\text{ function }n \neq 0$ $8n = 64 \log_2(n)$ $\dfrac{n}{8} = \log_2(n)$ $2^{n/8} = n$ 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?

March 5th, 2019, 12:53 PM   #4
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Quote:
 Originally Posted by babaliaris 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 $n\neq 0$

 March 5th, 2019, 12:55 PM #5 Senior Member   Joined: Dec 2015 From: iPhone Posts: 437 Thanks: 68 n cannot be zero . Before starting to solve an equation the domain must always be found first . Thanks from babaliaris

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