Evaluate this expression

chijioke

Junior Member
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Jul 27, 2022
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We have explained this to you repeatedly, such as what I said here:

It is not something to prove mathematically, but a standard convention, that when we write something like [imath]-x^y[/imath] the exponent applies only to [imath]x[/imath], not to [imath]-x[/imath]. For evidence, consider these sources:


Some of these mention that some calculators or programs follow a different rule, but the mathematical rule is as we say.

So your thinking is wrong. Rather than your [math]-2^6=(-2)^6=(-2)(-2)(-2)(-2)(-2)(-2)=64[/math]the reality is [math]-2^6=-(2^6)=-(2)(2)(2)(2)(2)(2)=-64[/math]
I am fully convinced now.
 

chijioke

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Jul 27, 2022
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100
[math]\frac{ 1 }{( - 8 ) ^ {( - \frac{ 1 }{ 3 })}} + \frac{ 1 }{( - 32 ) ^ {( - \frac{ 1 }{ 5 })}} - \frac{ 1 }{( - 64 ) ^ {( - \frac{ 1 }{ 3 })}} \\ = \left( ( - 8 ) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} + \left( ( - 32 ) ^ {( - \frac{ 1 }{ 5 })} \right) ^ {( - 1 )} - \left( ( - 64 ) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} \\ = ( - 8 ) ^ { \left( ( - \frac{ 1 }{ 3 }) \times ( - 1 ) \right) } + ( - 32 ) ^ { \left( ( - \frac{ 1 }{ 5 }) \times ( - 1 ) \right) } - ( - 64 ) ^ { \left( ( - \frac{ 1 }{ 3 }) \times ( - 1 ) \right) } \\ = ( - 8 ) ^ {\frac{ 1 }{ 3 }} + ( - 32 ) ^ {\frac{ 1 }{ 5 }} - ( - 64 ) ^ {\frac{ 1 }{ 3 }} \\ = ( - 2 ^ { 3 }) ^ {\frac{ 1 }{ 3 }} + ( - 2 ^ { 5 }) ^ {\frac{ 1 }{ 5 }} - \left( ( - 1 ) \times 2 ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }} \\ = ( - 2 ) + ( - 2 ) - \left( ( - 1(2) ^ { 2 } \right) \\ = ( - 2 ) - 2 - \left( - 1 (4) \right) \\ = - 2 - 2 - (-4) \\ = - 2 - 2 + 4 \\ = -4 + 4 \\ = 0[/math]
Or
[math]\frac{ 1 }{( - 8 ) ^ {( - \frac{ 1 }{ 3 })}} + \frac{ 1 }{( - 32 ) ^ {( - \frac{ 1 }{ 5 })}} - \frac{ 1 }{( - 64 ) ^ {( - \frac{ 1 }{ 3 })}} \\ = \left( ( - 8 ) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} + \left( ( - 32 ) ^ {( - \frac{ 1 }{ 5 })} \right) ^ {( - 1 )} - \left( ( - 64 ) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} \\ = \left( ( - 2 ^ { 3 }) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} + \left( ( - 2 ^ { 5 }) ^ {( - \frac{ 1 }{ 5 })} \right) ^ {( - 1 )} - \left( ( - 4 ^ { 3 }) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} \\ = ( - 2 ^ { 3 }) ^ {\frac{ 1 }{ 3 }} + ( - 2 ^ { 5 }) ^ {\frac{ 1 }{ 5 }} - ( - 4 ^ { 3 }) ^ {\frac{ 1 }{ 3 }} \\ = -2 + (-2) - (-4) \\ = -2-2+4 \\ = -4+4 \\ = 0[/math]
 

chijioke

Junior Member
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How do I check to see if the solution is correct? As you know it is possible to check if one work/solution is correct for instance take a look at this:
Simplify
[math]\begin{array}{cccc} a. & \frac{ P ^ {\frac{ 3 }{ 2 }} P ^ {( - \frac{ 3 }{ 4 })}}{ P ^ {( - \frac{ 5 }{ 4 })}} \\[8pt] b. & \frac{ x ^ {\frac{ 2 }{ 3 }} y ^ {( - \frac{ 5 }{ 4 })}}{ x ^ {( - 6 )} y ^ {( - \frac{ 2 }{ 3 })}} \\[8pt] c. & 27 ^ { x } = \left( \frac{ 1 }{ 3 } \times \sqrt{ 81 } \right) \\[8pt] d. & 8 ^ { x } = \left( 2\sqrt{ 32 } \right) \end{array}[/math]When I simplified problem a.
[math]\frac{ P ^ {\frac{ 3 }{ 2 }} P ^ {( - \frac{ 3 }{ 4 })}}{ P ^ {( - \frac{ 5 }{ 4 })}}[/math]I got \(\displaystyle P^2\)
I went further to check by substituting 5 for P in the solution and the original expression.
That is
[math]\frac{ 5 ^ {\frac{ 3 }{ 2 }} 5 ^ {( - \frac{ 3 }{ 4 })}}{ 5 ^ {( - \frac{ 5 }{ 4 })}}[/math]AND ALSO \(\displaystyle 5^2\)
The result, twenty-five is the same in each case showing that the solution is correct.
When I simplified b.
[math]\frac{ x ^ {\frac{ 2 }{ 3 }} y ^ {( - \frac{ 5 }{ 4 })}}{ x ^ {( - 6 )} y ^ {( - \frac{ 2 }{ 3 })}}[/math]I got \(\displaystyle \frac{ \sqrt[ 3 ]{ x ^ { 20 }} }{ \sqrt[ 12 ]{ y ^ { 7 }} }\)
l checked by substituting 5 for x and 4 for y respectively in the original and final expression. That is
[math]\frac{ 5 ^ {\frac{ 2 }{ 3 }} 4 ^ {( - \frac{ 5 }{ 4 })}}{ 5 ^ {( - 6 )} 4 ^ {( - \frac{ 2 }{ 3 })}}[/math]
AND ALSO \(\displaystyle \frac{ \sqrt[ 3 ]{ 5 ^ { 20 }} }{ \sqrt[ 12 ]{ 4 ^ { 7 }} }\)

The result (20351.5911) is the same.
I solve for x in
[math]27 ^ { x } = \left( \frac{ 1 }{ 3 } \sqrt{ 81 } \right)[/math]At the end of the solving x was \(\displaystyle \frac{1 }{ 3 }\)
I checked by putting \(\displaystyle \frac{ 1 }{ 3}\)
for x in the left side of the equation and evaluating each of the expression at the left and right side of the equation. I got 3. showing that the work/solution is correct.
That is
[math]\begin {array}{cc} 27 ^{ \frac{1}{3}}= \left( \frac{ 1 }{ 3 } \sqrt{ 81 } \right) \\[5pt] 3=3\end{array}[/math]When I solved d, the last one
[math]8 ^ { x } = \left( 2\sqrt{ 32 } \right)[/math]At the end of the solving, x was \(\displaystyle \frac{7 }{ 6 }\)
I checked by putting \(\displaystyle \frac{ 7 }{ 6 }\)
for x at the left side of the equation and then evaluating the left and right side of the equation and got 11.3137 which is same for the left and right side of the equation, proving that the solution is correct.
Back to my question supposing am alone evaulating this
\(\displaystyle \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}}\)
How do check if the solution is correct at the end of the evaluation just like in the instances I gave?
 

Dr.Peterson

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Nov 12, 2017
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14,450
Back to my question supposing am alone evaulating this
\(\displaystyle \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}}\)
How do check if the solution is correct at the end of the evaluation just like in the instances I gave?
Checking a simplification is different from checking a solution of an equation; but fortunately, this is a numerical expression, so the obvious thing is just to type the whole thing into an algebraic calculator, or equivalent.

But that's a little more troublesome than I expected. I first typed

1/(-8)^(-1/3)+1/(-32)^(-1/5)-1/(-64)^(-1/3)​

into Google, and got this:

1664315598046.png

Oops! It looks like they used complex roots. I've seen that happen before, there or on some calculators, because they handle exponents in funny ways.

Then I pasted the same expression into the Windows calculator, and got this:

1664315712021.png

I was confused for a while, but eventually realized that this is the value of just the last exponent I'd entered, because I didn't end it with "="; when I hit Enter after this, I got

1664316452938.png

This results from a rounding error somewhere. But it's interesting how it interpreted what I entered (for example, "-" is taken as subtraction, not negation, but it inserted 0 appropriately.

I tried again, entering everything by hand following the calculators rules (e.g. 1 divided by 8 negate ^ 3 negate reciprocal ...), and got the same final result.

Finally I pasted the same thing into Excel and got the right answer, 0. A physical calculator should do the same.

But these various errors are instructive!

Now, if you want to check without using technology, then the thing to do is to work it out again a different way.
 

chijioke

Junior Member
Joined
Jul 27, 2022
Messages
100
Checking a simplification is different from checking a solution of an equation; but fortunately, this is a numerical expression, so the obvious thing is just to type the whole thing into an algebraic calculator, or equivalent.
I thought the same initially, but the funny answers calculators were giving me made to come online and ask for help.


Oops! It looks like they used complex roots. I've seen that happen before, there or on some calculators, because they handle exponents in funny ways.
Could it be the same as my calculator gave me below?

[math]618.0339887 m - \left( 556.480303 m i \right)[/math]Though I checked other ways answer the could be given and finally I got zero.
 

Dr.Peterson

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Nov 12, 2017
Messages
14,450
I thought the same initially, but the funny answers calculators were giving me made to come online and ask for help.


Could it be the same as my calculator gave me below?

[math]618.0339887 m - \left( 556.480303 m i \right)[/math]Though I checked other ways answer the could be given and finally I got zero.
Do you know about the fact that every number has three cube roots, two of them non-real? In your work you were, as we usually would, taking the real roots; so a calculator that gives a complex answer is giving a different kind of answer than you need. You just have to find the right calculator.
 

Cubist

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Oct 29, 2019
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1,654
@chijioke you might be able to configure your calculator to stop giving complex answers (and you should be able to turn it back on too, when you need them). Have a look in your calculator's manual under, "complex numbers".

EDIT: In Matlab you can use the function "nthroot" as opposed to "^"
 

Dr.Peterson

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Nov 12, 2017
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14,450
How? Could you explain with example(s)?
Have you learned about complex numbers at all? If so, then here is one source I found that discusses this, with an example:


What I said before was a little too brief. Every (real or complex) number except zero has n nth roots (e.g. 3 cube roots); and every real number except has 3 cube roots, one of which is real and the other two not.

The calculators that gave complex answers use an algorithm that, instead of giving preference to the real root of a negative number, seem to choose instead the first complex root (lowest argument) in some cases. So Google says that (-8)^(1/3) = -2, but 1/((-8)^(-1/3)) = 1 + 1.73205081 i.
 
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