My Math Forum Help with Logarithms

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 April 21st, 2017, 07:29 AM #1 Newbie   Joined: Feb 2017 From: USA Posts: 22 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: 6,672 Thanks: 2171 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: 7,472 Thanks: 884 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: 7,472 Thanks: 884 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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