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I have two questions:

  1. I need to solve for x here

$$0.95 = \exp(-(1+0.4\frac{x-20}{4})^{-\frac{1}{0.4}})$$

My steps:

$$\ln(0.95) = -(1+0.4\frac{x-20}{4})^{-\frac{1}{0.4}}$$

$$\ln(0.95)^{-0.4} = -(1+0.4\frac{x-20}{4})$$

Then I punch in calculator $\ln(0.95)^{-0.4} = 3.2808$. Negating and subtracting $1$, multiplying by $4$ dividing by $0.4$ and add $20$, I get $-22.8077$. However, the right answer is $42.8077$.

I thought I performed all steps correctly, so where did I go wrong?

  1. A more general question regarding how exponents work. I know that:

$$a^{b} = \exp(\ln(a)*b)$$

And it makes sense that, $$(-5)^{0.5} = \exp(\ln(-5)*0.5)$$ is undefined.

However, when I punch in, $$(-5)^{0.4} $$ to my TI-30XS Multiview calculator, I got 1.903653939. No $i$ what so ever.

So I guess I don't know how exponents work any more.

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    $\begingroup$ You raised both sides to the $-0.4$ power before multiplying both sides by $-1$. You need to first multiply both sides by $-1$ and then raise both sides to the $-0.4$ power. $\endgroup$ – Mark Viola Jul 2 '18 at 22:13
  • $\begingroup$ @MarkViola Right. Now I see the mistake.... $\endgroup$ – user101998 Jul 2 '18 at 22:15
  • $\begingroup$ To answer your second question, $\log(-5)=\log(5)+i(2n+1)\pi$ is multivalued. For $n=2$, $2n+1=5$ and $0.4 \times 5\pi=2\pi$. And since $e^{i2\pi}=1$, there is a real value of $(-5)^{0.4}$. Your calculator returns that value. $\endgroup$ – Mark Viola Jul 2 '18 at 22:16
  • $\begingroup$ How is $n$ determined? Also, if I were to enter $(-5)^{0.3}$, the calculator returns a domain error. What is explaining this? $\endgroup$ – user101998 Jul 2 '18 at 22:34
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    $\begingroup$ $n$ can be any integer, positive or negative. There are no real numbers equal to $$(-0.5)^{0.3}=e^{0.3 \log(0.5)+i(2n+1)0.3}$$ $\endgroup$ – Mark Viola Jul 3 '18 at 0:37
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From here

$$\ln(0.95) = -\left(1+0.4\frac{x-20}{4}\right)^{-\frac{1}{0.4}}$$

we have

$$-\ln(0.95) = \left(1+0.4\frac{x-20}{4}\right)^{-\frac{1}{0.4}}$$

and then

$$[-\ln(0.95)]^{-0.4} = \left(1+0.4\frac{x-20}{4}\right)$$

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  • $\begingroup$ I believe the OP meant $[\ln(0.95)]^{-0.4}$. $\endgroup$ – N. F. Taussig Jul 2 '18 at 22:10
  • $\begingroup$ @N.F.Taussig Ops...I think you are right! Thanks $\endgroup$ – gimusi Jul 2 '18 at 22:11
  • $\begingroup$ Why do you think this response is better suited to an answer than to a comment? $\endgroup$ – Robert Howard Jul 2 '18 at 22:17
  • $\begingroup$ @RobertHoward I don't think so! Anyone is free to choose to give an answer or let a comment. I prefer to give an answer but I don't think this is better than a good comment. $\endgroup$ – gimusi Jul 2 '18 at 22:19
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Let's first simplify by setting $$ y=1+0.4\frac{x-20}{4}=1+\frac{x-20}{10}=\frac{x}{10}-1 $$ so that your expression becomes $$ 0.95=\exp(-y^{-1/0.4}) $$ Taking logarithms: $$ \ln0.95=-y^{-1/0.4} $$ Negate and raise both sides to $-0.4$: $$ (-\ln0.95)^{-0.4}=y $$ With a calculator, $$ y=3.280769831611707 $$ Therefore $$ x=10(y+1)=42.807698316117066 $$

Note that $\ln0.95<0$, so raising it to $-0.4$ is not really the best thing to do: better dealing with positive numbers.

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The negation needs to come before the exponentiation. This will turn your $-22.8076983$ into a positive $22.8076983$. Plus, you don't need to deal with complex numbers.

Have you ever seen the equation $e^{\pi i} = -1$? Have you ever noticed that exponentiation always results in a positive number when you're working with the "Real" number line? What happened was you were trying to invert an exponentiation with a logarithm, but since it's negative, that only makes sense in the "Complex" number system. Complex logarithms can end up having multiple answers ("branches"?), though I'm not sure if that's what you encountered there.

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