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March 5th, 2019, 01:20 AM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 139 Thanks: 8  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, 08:34 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 
$8n^2  64n \log_2(n) = 0$ $8n^2 = 64 n \log_2(n)$ $\text{as }n \text{ is the argument of the }\log_2()\text{ function }n \neq 0$ $8n = 64 \log_2(n)$ $\dfrac{n}{8} = \log_2(n)$ $2^{n/8} = n$ 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 $x \approx 1.099997,~43.55926$ 
March 5th, 2019, 12:02 PM  #3  
Senior Member Joined: Oct 2015 From: Greece Posts: 139 Thanks: 8  Quote:
 
March 5th, 2019, 01:53 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476  
March 5th, 2019, 01:55 PM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 834 Thanks: 113 Math Focus: Elementary Math  n cannot be zero . Before starting to solve an equation the domain must always be found first . 

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depented, equations, logarithmic, multiple, solve, terms 
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