Evaluate this expression

[chijioke]

\(\displaystyle \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}}\)

Solution​

[imath]\begin{array}{llllllll} \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}} \\[11pt] \newline = \frac{ 1 }{\frac{ 1 }{( - 8 ) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{( - 32 ) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{( - 64 ) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline = \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 3 } \right) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 5 } \right) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline =\frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{( - 2 ) ^ { 2 }} \\[8pt] = - 2 + (- 2 ) - ( - 2 ) ^ { 2 } \\ = - 2 - 2 - (+4) \\ = -4 - 4 = - 8 \end{array}[/imath]
Is my solution correct?
Am not sure if work is correct. Please help me check.

 

[Dr.Peterson]

\(\displaystyle \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}}\)

Solution​

[imath]\begin{array}{llllllll} \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}} \\[11pt] \newline = \frac{ 1 }{\frac{ 1 }{( - 8 ) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{( - 32 ) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{( - 64 ) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline = \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 3 } \right) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 5 } \right) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline =\frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{( - 2 ) ^ { 2 }} \\[8pt] = - 2 + (- 2 ) - ( - 2 ) ^ { 2 } \\ = - 2 - 2 - (+4) \\ = -4 - 4 = - 8 \end{array}[/imath]
Is my solution correct?
Am not sure if work is correct. Please help me check.

There's a problem with the third term. -64 is not equal to (-2)^6.

 

[chijioke]

There's a problem with the third term. -64 is not equal to (-2)^6.

Thanks for the observation.
\(\displaystyle \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}}\)

Solution​

[imath]\begin{array}{llllllllll} \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}} \\[11pt] \newline = \frac{ 1 }{\frac{ 1 }{( - 8 ) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{( - 32 ) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{( - 64 ) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline = \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 3 } \right) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 5 } \right) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{ \left( 2 ) ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \text{because} -2^6 = 64 ~ \text{and} \ne ~ -64 \\ =\frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{( 2 ) ^ { 2 }} \\[8pt] = \frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{ 4 } \\[7pt] = - 2 + (- 2 ) - ( 4) \\ = - 2 - 2 - 4 \\ = -4 - 4 = - 8 \end{array}[/imath]
Is it okay now?

 

[Dr.Peterson]

[imath]\begin{array}{llllllllll} \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}} \\[11pt] \newline = \frac{ 1 }{\frac{ 1 }{( - 8 ) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{( - 32 ) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{( - 64 ) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline = \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 3 } \right) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 5 } \right) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{ \left( 2 ) ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \text{because} -2^6 = 64 ~ \text{and} \ne ~ -64 \\ =\frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{( 2 ) ^ { 2 }} \\[8pt] = \frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{ 4 } \\[7pt] = - 2 + (- 2 ) - ( 4) \\ = - 2 - 2 - 4 \\ = -4 - 4 = - 8 \end{array}[/imath]
Is it okay now?

No. -64 is not equal to 2^6 either!

 

[topsquark]

\(\displaystyle \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}}\)

Solution​

[imath]\begin{array}{llllllll} \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}} \\[11pt] \newline = \frac{ 1 }{\frac{ 1 }{( - 8 ) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{( - 32 ) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{( - 64 ) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline = \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 3 } \right) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 5 } \right) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline =\frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{( - 2 ) ^ { 2 }} \\[8pt] = - 2 + (- 2 ) - ( - 2 ) ^ { 2 } \\ = - 2 - 2 - (+4) \\ = -4 - 4 = - 8 \end{array}[/imath]
Is my solution correct?
Am not sure if work is correct. Please help me check.

A suggestion that will help you out a bit.
[imath]x^{-1} = \dfrac{1}{x}[/imath]

But that means
[imath]\dfrac{1}{x^{-1}} = \left ( x^{-1} \right )^{-1} = x^{(-1)(-1)} = x^1 = x[/imath].

So a negative exponent in the denominator flips the number to the numerator. There is no need to write something like
[imath]\dfrac{1}{x^{-1}} = \dfrac{ 1 }{ \left ( \dfrac{1}{x} \right ) }[/imath]

-Dan

 

[Steven G]

Just like + tells you to add, the negative sign in the exponent tells you to take the reciprocal.

 

[Steven G]

(-64)^1/3 = -4 since (-4)(-4)(-4) = -64

 

[Harry_the_cat]

Think of \(\displaystyle 64\) as \(\displaystyle 4^3\) rather than \(\displaystyle 2^6\).

 

[chijioke]

No. -64 is not equal to 2^6 either!

I mean \(\displaystyle -2^6=64\)
because \(\displaystyle - 2 \times - 2 \times - 2 \times - 2 \times - 2 \times - 2 = 64 \)

 

[Steven G]

Sorry, but -2⁶ is not 64
How much is 100-2⁶? Is it 36 or 164?
How much is 0-2⁶?
How much is 2⁶?
How much is -2⁶?
What do you get if you put a negative sign in front of 64?

 

[Dr.Peterson]

[imath]\begin{array}{llllllllll} \frac{ 1 }{( - 8 ) ^ {\frac{ - 1 }{ 3 }}} + \frac{ 1 }{( - 32 ) ^ {\frac{ - 1 }{ 5 }}} - \frac{ 1 }{( - 64 ) ^ {\frac{ - 1 }{ 3 }}} \\[11pt] \newline = \frac{ 1 }{\frac{ 1 }{( - 8 ) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{( - 32 ) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{( - 64 ) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \newline = \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 3 } \right) ^ {\frac{ 1 }{ 3 }}}} + \frac{ 1 }{\frac{ 1 }{ \left( ( - 2 ) ^ { 5 } \right) ^ {\frac{ 1 }{ 5 }}}} - \frac{ 1 }{\frac{ 1 }{ \left( 2 ) ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }}}} \\[14pt] \text{because} -2^6 = 64 ~ \text{and} \ne ~ -64 \\ =\frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{( 2 ) ^ { 2 }} \\[8pt] = \frac{\frac{ 1 }{ 1 }}{ - 2 } + \frac{\frac{ 1 }{ 1 }}{ - 2 } - \frac{\frac{ 1 }{ 1 }}{ 4 } \\[7pt] = - 2 + (- 2 ) - ( 4) \\ = - 2 - 2 - 4 \\ = -4 - 4 = - 8 \end{array}[/imath]
Is it okay now?

I mean \(\displaystyle -2^6=64\)
because \(\displaystyle - 2 \times - 2 \times - 2 \times - 2 \times - 2 \times - 2 = 64 \)

My comment was about your replacing [imath](-64)^\frac{1}{3}[/imath] with [imath](2^6)^\frac{1}{3}[/imath]. Your reply is irrelevant. It is not true that [imath]-64=2^6[/imath]. You simply dropped a negative sign, and (if you go the way you are going, which is extremely inefficient) the right thing to have done is to change [imath](-64)^\frac{1}{3}[/imath] to [imath](-2^6)^\frac{1}{3}[/imath].

As for \(\displaystyle -2^6=64\), that is wrong because the order of operations tells us to do the exponent before the negation: \(\displaystyle -2^6=-(2^6)=-64\). It does not mean [imath](-2)^6[/imath]; it is not -2, but 2 that is raised to the power.

In my opinion, it's time to show you how to apply some of the hints that have been given. Here is how I might start the work:

[math]\frac{1}{(-8)^{-\frac{1}{3}}}+\frac{1}{(-32)^{-\frac{1}{5}}}-\frac{1}{(-64)^{-\frac{1}{3}}}=\\(-8)^{\frac{1}{3}}+(-32)^{\frac{1}{5}}-(-64)^{\frac{1}{3}}=\\((-2)^3)^{\frac{1}{3}}+((-2)^5)^{\frac{1}{5}}-((-4)^3)^{\frac{1}{3}}[/math]
Do you follow that? If so, continue.

 

[Harry_the_cat]

I mean \(\displaystyle -2^6=64\)
because \(\displaystyle - 2 \times - 2 \times - 2 \times - 2 \times - 2 \times - 2 = 64 \)

Note the difference between
\(\displaystyle -2^6 = - 2\times2\times2\times2\times2\times2 = -64\)
and
\(\displaystyle (-2)^6 = -2 \times-2\times-2\times-2\times-2\times-2 = +64\)

 

[chijioke]

Sorry, but -2⁶ is not 64
How much is 100-2⁶? Is it 36 or 164?
How much is 0-2⁶?
How much is 2⁶?
How much is -2⁶?
What do you get if you put a negative sign in front of 64?

Think of \(\displaystyle 64\) as \(\displaystyle 4^3\) rather than \(\displaystyle 2^6\).

Note the difference between
\(\displaystyle -2^6 = - 2\times2\times2\times2\times2\times2 = -64\)
and
\(\displaystyle (-2)^6 = -2 \times-2\times-2\times-2\times-2\times-2 = +64\)

My comment was about your replacing [imath](-64)^\frac{1}{3}[/imath] with [imath](2^6)^\frac{1}{3}[/imath]. Your reply is irrelevant. It is not true that [imath]-64=2^6[/imath]. You simply dropped a negative sign, and (if you go the way you are going, which is extremely inefficient) the right thing to have done is to change [imath](-64)^\frac{1}{3}[/imath] to [imath](-2^6)^\frac{1}{3}[/imath].

As for \(\displaystyle -2^6=64\), that is wrong because the order of operations tells us to do the exponent before the negation: \(\displaystyle -2^6=-(2^6)=-64\). It does not mean [imath](-2)^6[/imath]; it is not -2, but 2 that is raised to the power.

In my opinion, it's time to show you how to apply some of the hints that have been given. Here is how I might start the work:

[math]\frac{1}{(-8)^{-\frac{1}{3}}}+\frac{1}{(-32)^{-\frac{1}{5}}}-\frac{1}{(-64)^{-\frac{1}{3}}}=\\(-8)^{\frac{1}{3}}+(-32)^{\frac{1}{5}}-(-64)^{\frac{1}{3}}=\\((-2)^3)^{\frac{1}{3}}+((-2)^5)^{\frac{1}{5}}-((-4)^3)^{\frac{1}{3}}[/math]
Do you follow that? If so, continue.

All the replies noted. But I think I would prefer to use \(\displaystyle 2^6\)

[math] \frac{ 1 }{( - 8 ) ^ {( - \frac{ 1 }{ 3 })}} + \frac{ 1 }{( - 32 ) ^ {( - \frac{ 1 }{ 5 })}} - \frac{ 1 }{( - 64 ) ^ {( - \frac{ 1 }{ 3 })}} \\[5pt] = \left( ( - 8 ) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} + \left( ( - 32 ) ^ {( - \frac{ 1 }{ 5 })} \right) ^ {( - 1 )} - \left( ( - 64 ) ^ {( - \frac{ 1 }{ 3 })} \right) ^ {( - 1 )} \\[5pt] = ( - 8 ) ^ { \left( ( - \frac{ 1 }{ 3 }) \times ( - 1 ) \right) } + ( - 32 ) ^ { \left( ( - \frac{ 1 }{ 5 }) \times ( - 1 ) \right) } - ( - 64 ) ^ { \left( ( - \frac{ 1 }{ 3 }) \times ( - 1 ) \right) } \\[5pt] = ( - 8 ) ^ {\frac{ 1 }{ 3 }} + ( - 32 ) ^ {\frac{ 1 }{ 5 }} - ( - 64 ) ^ {\frac{ 1 }{ 3 }} \\[4pt] = ( - 2 ^ { 3 }) ^ {\frac{ 1 }{ 3 }} + ( - 2 ^ { 5 }) ^ {\frac{ 1 }{ 5 }} - \left( ( - 1 ) \times 2 ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }} \\[4pt] = ( - 2 ) + ( - 2 ) - \left( ( - 1(2) ^ { 2 } \right) \\[3pt] = ( - 2 ) - 2 - \left( - 1 (4) \right) \\[2pt] = - 2 - 2 - (-4) // = - 2 - 2 + 4 \\ = -4 + 4 \\ = 0 [/math]Is it okay now?

 

[Steven G]

You have the correct answer but I'm not sure that you just weren't lucky.
Here are my concerns:
1) You had ((-1)x2⁶)^1/3 and you wrote that equals (-1)(2)². My question is did you write -1 because you computed that (-1)^1/3= -1 or did you just ignore that -1 was being raised to the 1/3 power?

2) Did you replace (-8)with -2³ because you know that (-8)=(-2)³=-2³?
If the power would have been even, say 4, then for example 16 = (-2)⁴ = -2⁴=-16 is wrong! Which equal sign is incorrect.

 

[chijioke]

You have the correct answer but I'm not sure that you just weren't lucky.
Here are my concerns:
1) You had ((-1)x2⁶)^1/3 and you wrote that equals (-1)(2)². My question is did you write -1 because you computed that (-1)^1/3= -1 or did you just ignore that -1 was being raised to the 1/3 power?

I think I made an error there. It should have been something like this
[math] = ( - 2 ^ { 3 }) ^ {\frac{ 1 }{ 3 }} + ( - 2 ^ { 5 }) ^ {\frac{ 1 }{ 5 }} - \left( ( - 1 ) \left( 2 ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }} \right)\\[3pt] = ( - 2 ) + ( - 2 ) - \left( ( - 1)(2) ^ { 2 } \right) \\[3pt] = ( - 2 ) - 2 - \left( - 1 (4) \right) \\[2pt] = - 2 - 2 - (-4) \\ = - 2 - 2 + 4 \\ = -4 + 4 \\ = 0 [/math]This is because if it appears like this
[math] \left( - 1 \times 2 ^ { 6 } \right) ^ {\frac{ 1 }{ 3 }} [/math]It means am forgetting too quickly that \(\displaystyle -1 \) is also being raised to power [imath]\frac{1}{3}[/imath]
which is not supposed to be so if \(\displaystyle -64\) must maintain its value as \(\displaystyle -64\text{~.}\)

2) Did you replace (-8)with -2³ because you know that (-8)=(-2)³=-2³?

Ya. I know because [math](-2) × (-2) × (-2)~\text{still gives me}~ - 8[/math]

If the power would have been even, say 4, then for example 16 = (-2)⁴ = -2⁴=-16 is wrong! Which equal sign is incorrect.

Either sign could be used since
[math](-2)×(-2)×(-2)×(-2) = 16\\ \text{and also}~(2)×(2)×(2)×(2)=16[/math]Even power always gives positive sign.

 

[chijioke]

Note the difference between
\(\displaystyle -2^6 = - 2\times2\times2\times2\times2\times2 = -64\)

Why did you put the -2 only on the first 2?

and
\(\displaystyle (-2)^6 = -2 \times-2\times-2\times-2\times-2\times-2 = +64\)

I stand to be corrected please. Is
[math]-2^6 \ne(-2)^6[/math]I am thinking that
[math]-2^6=(-2)^6=64[/math]If you say that [math]-2^6=-2×2×2×2×2×2 \\ \text{instead of}~ (-2)×(-2)×(-2)×(-2)×(-2)×(-2)[/math]Could you please prove it?

 

[Dr.Peterson]

Why did you put the -2 only on the first 2?

I stand to be corrected please. Is
[math]-2^6 \ne(-2)^6[/math]I am thinking that
[math]-2^6=(-2)^6=64[/math]If you say that [math]-2^6=-2×2×2×2×2×2 \\ \text{instead of}~ (-2)×(-2)×(-2)×(-2)×(-2)×(-2)[/math]Could you please prove it?

We have explained this to you repeatedly, such as what I said here:

As for \(\displaystyle -2^6=64\), that is wrong because the order of operations tells us to do the exponent before the negation: \(\displaystyle -2^6=-(2^6)=-64\). It does not mean [imath](-2)^6[/imath]; it is not -2, but 2 that is raised to the power.

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:

What is the rule for negative numbers squared?

Squaring a negative number multiplies it by itself, meaning two minus signs that cancel; e.g. (−3)² = 3² = 9. The rule is to cancel pairs of minus signs.

www.purplemath.com

Order of operations - Wikipedia

en.wikipedia.org

Order of Operations: Common Misunderstandings – The Math Doctors

www.themathdoctors.org

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]

 

[Steven G]

Why did you put the -2 only on the first 2 in -2^6?

I asked you to answer these questions before.
How much is 100-2^6?
How much is 0-2^6?
How much is -2^6?

What do you get when you put a negative sign in front of 64?
How much is 2^6?
Using your answers to the last two questions above, how much is -2^6.


How much is (-2)^6?

 

[Harry_the_cat]

Why did you put the -2 only on the first 2? There is only one negative sign. The positive number \(\displaystyle 2\) is being raised to the sixth power. I didn't put the negative sign on the first \(\displaystyle 2\), I put it at the beginning of the expression.

I stand to be corrected please. Is
[math]-2^6 \ne(-2)^6[/math] YES that statement is correct ie they are NOT equal.

I am thinking that
[math]-2^6=(-2)^6=64[/math] NO In \(\displaystyle -2^6\), only the \(\displaystyle 2\) is being raised to the sixth power. If you want all of \(\displaystyle -2\) raised to the sixth power you need to put \(\displaystyle -2\) in brackets, so you get \(\displaystyle (-2)^6\).

If you say that [math]-2^6=-2×2×2×2×2×2 \\ \text{instead of}~ (-2)×(-2)×(-2)×(-2)×(-2)×(-2)[/math]Could you please prove it? There is nothing to prove. It is a convention.

See comments in red.

 

[chijioke]

I asked you to answer these questions before.
How much is 100-2^6?

[math]100-2^6 \\ =100 -2×2×2×2×2×2 \\ = 100 - 64 \\ = 36[/math]

How much is 0-2^6?

[math]0-2^6 \\ = 0 - 2×2×2×2×2×2 \\ = 0 - 64 \\ = -64[/math]

How much is -2^6?

[math]-2^6 \\ = 2×2×2×2×2×2 \\ = - 64[/math]

What do you get when you put a negative sign in front of 64?

I get -64.

How much is 2^6?

It is 64 because [math]2^6 \\ = 2×2×2×2×2×2 \\ = 64[/math]

Using your answers to the last two questions above, how much is -2^6.

[math]-2^6 = -64[/math]

How much is (-2)^6?

[math](-2)^6 = 64[/math]I am now convinced that
[math]-2^6 = -64[/math]and that
[math](-2)^6 = 64[/math]The two expressions are different with different meanings.