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April 21st, 2017, 07:29 AM  #1 
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1  Help with Logarithms
Hello all, When solving for a log problem, why don't we include negative answers? i.e. $\displaystyle log_{36}$x=2 Solving would require us to take the square root of 36 which could be either 6 or 6, no? I recall solving square root equations with a +/. Thanks in advance. 
April 21st, 2017, 07:46 AM  #2  
Senior Member Joined: May 2016 From: USA Posts: 919 Thanks: 368  Quote:
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First off $log_{36}(x) = 2 \implies 36^2 = x \implies x = 1296.$ Second, we only want positive answers for the argument of a logarithm. So the negative root can be ignored. Let's take this as an example. $log_{36}(y) = \dfrac{1}{2} \implies y = 36^{(1/2)} = \sqrt{36} = 6.$ We ignore $\ \sqrt{36}$ because we care only about positive answers in this case.  
April 21st, 2017, 08:49 AM  #3  
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1  Quote:
 
April 21st, 2017, 08:56 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,228 Thanks: 2411 Math Focus: Mainly analysis and algebra 
The base of a logarithm must be positive because $$\log_b x = y \iff x = b^y$$ But the function $b^y$ is not "well behaved" at all. It is undefined for many rational values of $y$ and for others it veers wildly between positive and negative values. All of this makes it difficult to define sensible values for irrational values of $y$. In practice, then, having $b \lt 0$ is totally useless. 
April 21st, 2017, 09:49 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,762 Thanks: 1010 Math Focus: Elementary mathematics and beyond 
$$e^{\pi i}=1$$

April 21st, 2017, 11:53 AM  #6 
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1  
April 21st, 2017, 12:03 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,762 Thanks: 1010 Math Focus: Elementary mathematics and beyond 
Yes, so $\log(1)=k\pi i$, where $k$ is an odd integer. The logarithm is base $e$.


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