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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
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Quote:
 Originally Posted by chopnhack Hello all, When solving for a log problem, why don't we include negative answers?
Logarithms are defined only for positive numbers.

Quote:
 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 +/-.
No.

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
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 Originally Posted by JeffM1 Logarithms are defined only for positive numbers. No. 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.
sorry about that, I was having trouble with the latex and didn't double check my work - I wanted logx36=2 not the other way around, thanks again! April 21st, 2017, 08:56 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 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. Thanks from chopnhack April 21st, 2017, 09:49 AM #5 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $$e^{\pi i}=-1$$ Thanks from chopnhack April 21st, 2017, 11:53 AM   #6
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 Originally Posted by greg1313 $$e^{\pi i}=-1$$
I like what you did there, LOL

euler's number raised to pi times imaginary number?  April 21st, 2017, 12:03 PM #7 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Yes, so $\log(-1)=k\pi i$, where $k$ is an odd integer. The logarithm is base $e$. Tags logarithms 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 a2910 Algebra 3 October 27th, 2014 03:14 PM ron246 Algebra 5 May 22nd, 2014 06:59 AM Mr Davis 97 Algebra 1 May 3rd, 2014 06:34 PM Mindless_1 Algebra 2 November 14th, 2012 12:02 PM jwood Algebra 3 March 25th, 2011 01:41 PM

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