Mathematics, 07.11.2020 08:30 tckryoung

Is it possible for a logarithm to equal a negative number and why?

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The only numbers you can plug into a logarithm are positive numbers not equal to 1. Negative numbers, and the number 0, aren’t acceptable arguments to plug into a logarithm, but why?

The reason has more to do with the base of the logarithm than with the argument of the logarithm. To understand why, we have to understand that logarithms are actually exponents. The base of a logarithm is also the base of a power function.

When you have a power function with base 0, the result of that power function is always going to be 0. In other words, there’s no exponent you can put on 0 that won’t give you back a value of 0. Or, put a different way, 0 raised to anything is always still 0. In the same way, 1 raised to anything is always still 1.

If you raise a negative number to a positive number that’s not an integer, but instead a fraction or a decimal, you might end up with a negative number underneath a square root. And as you know, unless we’re getting into imaginary numbers, we can’t deal with a negative number underneath a square root.

So 0, 1, and every negative number presents a potential problem as the base of a power function. And if those numbers can’t reliably be the base of a power function, then they also can’t reliably be the base of a logarithm.

For that reason, we only allow positive numbers other than 1 as the base of the logarithm. Then what we know is that, if the base of our power function is positive, it doesn’t matter what exponent we put on that base (it could be a positive number, a negative number, of 0), that power function is going to come out as a positive number.

So in summary, because the base can only be a positive number, that means the argument of the logarithm can only be a positive number. Which means that in order to protect our bases, we have to only allow positive arguments inside the logarithm.

Hope this helps!

Answer from: Quest

you should pay $0.88

what is the image?

:if cheese is $4.40 per kilogramme, what should i pay for 200 grammes?

Answer from: Quest

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

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