Question 390986
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So far, so good.


Now, remember that a negative exponent in a numerator translates to a positive exponent in the denominator.  That is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^{-1}\ =\ \frac{1}{x}]


or, more to the point:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v^{-2}\ =\ \frac{1}{v^2}]


so your equation becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{v^2}\ =\ \frac{1}{100}]


which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v^2\ =\ 100]


from which we see that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v\ =\ \pm10]


However, since in *[tex \Large \log_b(x)\ =\ y] we know that *[tex \Large b\ >\ 0] and *[tex \Large b\ \neq\ 1], hence we exclude *[tex \Large v\ =\ -10].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ v\ =\ 10]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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