- #1

- 97

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Say, log

_{-10}(100) = 2. Rewritten, -10

^{2}= 100, which is accurate.

You could suggest that you may as well just ignore the negative because -x

^{2}= x

^{2}, but it's still weird that this shows up as a syntax error on a calculator.

- Thread starter Rumplestiltskin
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- #1

- 97

- 3

Say, log

You could suggest that you may as well just ignore the negative because -x

- #2

mfb

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Or, alternatively, solve log

For integers (with the right sign) as logarithm values, it would work, but only for those.

- #3

Mark44

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No, it isn't. ##-10^2 = - 100## because the exponent has a higher precedence than the negation sign. What you probably meant was ##(-10)^2##, which is 100.I understand that taking logs of a negative number isn't possible because no number to any power produces a negative number. But why not a negative base?

Say, log_{-10}(100) = 2. Rewritten, -10^{2}= 100, which is accurate.

For the reason given above, ##-x^2 \ne x^2##, unless x happens to be 0. If you want to square a negative number on a calculator, put parentheses around the number, with the exponent outside the parentheses.Rumplestiltskin said:You could suggest that you may as well just ignore the negative because -x^{2}= x^{2}, but it's still weird that this shows up as a syntax error on a calculator.

- #4

Mark44

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- #5

SammyS

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You should be more careful regarding order of operations.

Say, log_{-10}(100) = 2. Rewritten, -10^{2}= 100, which is accurate.

You could suggest that you may as well just ignore the negative because -x^{2}= x^{2}, but it's still weird that this shows up as a syntax error on a calculator.

-10

and

-x

I assume you meant to write

(-10)

and

(-x)

The main problem with having a negative base in a logarithm, is that there is a problem defining a real valued exponential function having a negative base.

- #6

- 97

- 3

Syntax error on calculator. When typed into google, (-10)_{-10}(50)=x.

Or, alternatively, solve log_{-10}(x)=1.7. What is x=(-10)^{1.7}?

For integers (with the right sign) as logarithm values, it would work, but only for those.

Could you elaborate?The main problem with having a negative base in a logarithm, is that there is a problem defining a real valued exponential function having a negative base.

- #7

SammyS

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That's a complex number. Isn't that a problem for a real function?Syntax error on calculator. When typed into google, (-10)^{1.7}= 29.4590465 - 40.5468989 i. Woah. Still at a loss.

Could you elaborate?

Try (-10)

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