How to solve the equation x^-1.5=1

[gegogabanana]

Hi I'm a University Utrecht student studying micro biology so not the best at math, I'm trying to solve the equation in the title for fun but I can't really make sense of the explenations online. Some help would be lovely. Thanks in advance!

 

[donal1353]

When dealing with indices, 1 to the power of any thing will always equal 1. Therefore, in this equation, x is equal to 1.

In a more genral sence, you want to raise the left hand side to be equal to 1. Solving this equation would be as follows.

x^-1.5 = 1
(x^-1.5)^-2/3 = 1^-2/3
x = 1^-2/3
x = 1

Here we make use of the rule of indices that states: (a^p)^q) = a^(p*q)

Rembering that (-1.5 * -2/3) = 1
and that x^1 = x
and that 1 to the power of anything = 1

Hope this helps, I can explain further if you would like.

 

[The Highlander]

Hi I'm a University Utrecht student studying micro biology so not the best at math, I'm trying to solve the equation in the title for fun but I can't really make sense of the explenations online. Some help would be lovely. Thanks in advance!

We can't help you understand any "explenation" (sic) you have seen if we don't know what it is?

You might share a link to one (or more) of the "online" explanations you saw that you don't understand (and say what it is about it/them that puzzles you).

In the meantime, below is a graph of y = x^-1.5 and you can see that when x = 1 then x^-1.5 also equals 1. Therefore, the solution to the equation: x^-1.5 = 1 is: x = 1. Does that help?

​

You can view the original graph (and 'play around' with it if you wish; eg: zoom in or out/move it around) here.

Hope that helps.

 

[gegogabanana]

Thanks alot, I was wondering if it also had anything to do with complex numbers since you could rewrite the equation as the xth root of -1.5=1
but the solution 1 seems helpfull aswell!

 

[Dr.Peterson]

Thanks alot, I was wondering if it also had anything to do with complex numbers since you could rewrite the equation as the xth root of -1.5=1
but the solution 1 seems helpfull aswell!

You can get complex solutions if that's what you want; you'll be doing essentially the same thing as for the real case: raise both sides of the equation to the reciprocal power.

But x^-1.5 is not equivalent to the xth root of -1.5. The latter would be (-1.5)^(1/x).

 

[khansaheb]

Hi I'm a University Utrecht student studying micro biology so not the best at math, I'm trying to solve the equation in the title for fun but I can't really make sense of the explenations online. Some help would be lovely. Thanks in advance!

Thanks alot, I was wondering if it also had anything to do with complex numbers since you coul{d rewrite the equation as the xth root of -1.5=1
but the solution 1 seems helpfull aswell!

If you want to use complex variable, you can start with:

\(\displaystyle x^{-1.5} \ = 1 \)

\(\displaystyle x^{-1.5} \ = sin({{\pi}/2+2*n*\pi}) + i * cos({{\pi}/2+2*n*\pi})\)

\(\displaystyle x \ = \ [sin({{\pi}/2+2*n*\pi}) + i * cos({{\pi}/2+2*n*\pi})]^{{-1/{1.5}}} \)

continue......

 

[Dr.Peterson]

If you want to use complex variable, you can start with:

\(\displaystyle x^{-1.5} \ = 1 \)

\(\displaystyle x^{-1.5} \ = sin({{\pi}/2+2*n*\pi}) + i * cos({{\pi}/2+2*n*\pi})\)

\(\displaystyle x \ = \ [sin({{\pi}/2+2*n*\pi}) + i * cos({{\pi}/2+2*n*\pi})]^{{-1/{1.5}}} \)

continue......

That would be if the RHS were i rather than 1. As it is, you need

\(\displaystyle x^{-3/2} = 1 \)

\(\displaystyle x^{-3/2} = \sin(2n\pi) + i \cos(2n\pi)\)

\(\displaystyle x = \left [\sin(2n\pi) + i \cos(2n\pi)\right]^{{-2/3}} \)

 

[khansaheb]

That would be if the RHS were i rather than 1. As it is, you need

\(\displaystyle x^{-3/2} = 1 \)

\(\displaystyle x^{-3/2} = \sin(2n\pi) + i \cos(2n\pi)\)

\(\displaystyle x = \left [\sin(2n\pi) + i \cos(2n\pi)\right]^{{-2/3}} \)

I find

\(\displaystyle \sin(2n\pi) = 0 \)

\(\displaystyle \cos(2n\pi) = 1\)..........Thus

\(\displaystyle i = \sin(2n\pi) + i \cos(2n\pi)\)........................ not "1" as claimed in response #6

Please check my work.

 

[Dr.Peterson]

I find

\(\displaystyle \sin(2n\pi) = 0 \)

\(\displaystyle \cos(2n\pi) = 1\)..........Thus

\(\displaystyle i = \sin(2n\pi) + i \cos(2n\pi)\)........................ not "1" as claimed in response #6

Please check my work.

Yes, I missed the other difference from what I expected, namely that you swapped cos and sin from their usual order. So what I meant to say was this:

\(\displaystyle x^{-3/2} = 1 \)

\(\displaystyle x^{-3/2} = \cos(2n\pi) + i \sin(2n\pi)\)

\(\displaystyle x = \left [\cos(2n\pi) + i \sin(2n\pi)\right]^{{-2/3}} \)

And that is equivalent to what you originally wrote. I was reading it as "cis", and should have typed it in myself rather than editing yours, which would have helped me say what I mean (and maybe catch that your error wasn't really an error).

But is there a reason you chose your unusual formulation?? Was it a trap for people like me? Or just a way to "complexify" the work in two ways at once?

 

[khansaheb]

Yes, I missed the other difference from what I expected, namely that you swapped cos and sin from their usual order. So what I meant to say was this:

\(\displaystyle x^{-3/2} = 1 \)

\(\displaystyle x^{-3/2} = \cos(2n\pi) + i \sin(2n\pi)\)

\(\displaystyle x = \left [\cos(2n\pi) + i \sin(2n\pi)\right]^{{-2/3}} \)

And that is equivalent to what you originally wrote. I was reading it as "cis", and should have typed it in myself rather than editing yours, which would have helped me say what I mean (and maybe catch that your error wasn't really an error).

But is there a reason you chose your unusual formulation?? Was it a trap for people like me? Or just a way to "complexify" the work in two ways at once?

Actually, when I wrote the first line of response #5, I did not realize that I wrote "sic" instead of "cis". When I did - I was too lazy to use LaTex and fix it. I intended to fix it after OP's response