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April 21st, 2017, 08:29 AM   #1
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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.
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April 21st, 2017, 08:46 AM   #2
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Quote:
Originally Posted by chopnhack View Post
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.
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April 21st, 2017, 09:49 AM   #3
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Originally Posted by JeffM1 View Post
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!
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April 21st, 2017, 09:56 AM   #4
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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.
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April 21st, 2017, 10:49 AM   #5
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$$e^{\pi i}=-1$$
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April 21st, 2017, 12:53 PM   #6
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$$e^{\pi i}=-1$$
I like what you did there, LOL

euler's number raised to pi times imaginary number?
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April 21st, 2017, 01:03 PM   #7
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Yes, so $\log(-1)=k\pi i$, where $k$ is an odd integer. The logarithm is base $e$.
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