Question

Solve
\[log_3(-3)+log_3(-1)\]

Answer 1

I don't know how to do it because we didn't learn the base change formula in class but I want to know how to use it... so can you all work this one for me please...

Answer 2

1. Not an equation.
2. logarithm of a negative # is non-real
(aka. an imaginary number, and yes that's really what it's called)

Answer 3

But does \[log_3(-3*-1)\] make it positive...

Answer 4

Are you dealing with real numbers only or complex numbers? If that question doesn't make sense to you, the answer is you are dealing with real numbers.

That being the case, the expression doesn't make sense, because
\[ \log_3 x \]
does not exist for any value of \(x \leq 0 \).

Answer 5

Okay this is the original question \[log_3(x+3)+log_3(x+5)=1\]
And for my solutions I got x=-6 and x=-2 but I have to check but I don't know how to since I have not learned the base change formula is someone willing to show me how to check?

Answer 6

Well, x = -6 cannot be an answer for the reasons we just discussed.

x=-2 however is an answer because
log_3(1) = 0, as log_a(1) = 0 for every base a > 0.

and
log_3(3) = 1, as log_a(a) = 1 for every base a > 0.

Make sense?

Answer 7

Yes, it does, but why couldn't I did what I did before, that is, multiply them to make it positive?

Answer 8

Because log_3(-3) and log_3(-1) are meaningless expressions, in the realm of real numbers.

Just because you can do something with them algebraically which makes them meaningful doesn't mean the original expressions are.

Answer 9

For example, if we manipulated things algebraically despite the fact they were nonsense, we could do something like this:
\[ \frac{2x}{x^2 - x^2} - \frac{1}{x-x} = \frac{1}{x-x} + \frac{1}{x+x} - \frac{1}{x-x} \]
\[ = \frac{1}{2x} \]
which in turn implies
\[ (2x)^2 = x^2 - x^2 = 0 \ \ \ \hbox{ for all } x \]
which is just more nonsense.

Answer 10

@JamesJ did a much better job of explaining this I think. It reminds me of the flawed proof that 1+1=2 which fails to take in account that you cannot divide by zero. Some functions, when out of their domain simply do not exist or are undefined (although I always forget what the precise mathematician-prefered definitions of those terms are).

Answer 11

So, @JamesJ am just not supposed to take the log of a negative number regardless of if I can use a law to multiply it and make it positive...

Answer 12

Yes, that's right.

For every positive number \( a > 0 \), the function
\[ f(x) = \log_a x \]
is defined if and only if \( x > 0 \).

This makes complete sense using the intuitive definition of log. Log gives you the exponent of a which allows you to recover x. I.e., if
\[ x = a^y \]
then by definition of log, \( y = \log_a x \).

Hence if a is positive to begin with, how can it ever be the case that \( a^y \) is zero or negative?

Answer 13

It cannot. If a > 0, then \( a^y > 0 \) for every real number y.

Answer 14

@purplec16 in the specific definitions for that rule of logarithms, you actually cannot use it if it's out of the domain. I had to look it up in my calculus textbook for the restrictions, but it's there alright. How about them apples? ;D

Answer 15

I still don't understand, sorry...

Answer 16

Let's use a specific base, base 10.

What is log(10) = ?

Answer 17

What is the domain of a log?

Answer 18

1

Answer 19

and what is log(1/10) = ?

Answer 20

-1

Answer 21

and log(10^(-100)) ?

Answer 22

The rule of logarithms you were using to combine, they have special restrictions.

The domain for logarithms are:
(0,\(\infty\)) i.e.: doesn't include zero

Answer 23

error

Answer 24

No,
\[ \log_{10} 10^{-100} = -100 \]

Answer 25

yes?

Answer 26

Yes :)

Answer 27

Now, what is log(0)?

Answer 28

log(0.001) = -3

Because one thousandth is 10^(-3) :-)

Answer 29

error...

Answer 30

right, not defined. Why? Because there is no real number x such that
\[ 10^x = 0 \]

Answer 31

Similarly, log(-1) is also not defined, because there is no real number x such that
\[ 10^x = -1 \]

Answer 32

Try ln(0) as well :D (that's a natural log)

Any # raised to the 0th power is 1.

Even this crazy, mind-blowing expression: 0\(^0\)

Answer 33

Okay, thank you!!! I think it's more clear!!! Thanks for all your help!

Answer 34

In fact, for every negative number, call it c < 0,
there is no real number x such that
\[ 10^x = c \]
Therefore log only accepts positive numbers as arguments.

Answer 35

Why? Because mathematicians said so. No really, I'm not kidding.

Answer 36

This is where your definitions and restrictions become important

Answer 37

Hey you're welcome @purplec16 , be sure to spread the knowledge around sometime eh? (although I wouldn't recommend bringing this up at the sports bar)

Answer 38

Lol! Thanks so much for your help once again, and why not at a sports bar @agentx5