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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 $x \approx 1.099997,~43.55926$ Thanks from topsquark and idontknow March 5th, 2019, 12:02 PM   #3
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 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, 01:53 PM   #4
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 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, 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 . Thanks from babaliaris Tags depented, equations, logarithmic, multiple, solve, terms Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathkid Calculus 6 November 29th, 2012 05:40 AM mathstud Algebra 33 July 23rd, 2012 12:19 PM hannah2329 Algebra 6 November 4th, 2011 02:38 AM philip Algebra 3 December 30th, 2010 08:57 AM philip Calculus 3 December 31st, 1969 04:00 PM

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