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Mathematics 13 Online
niahmk:

Simplify (3^4)−1. one over three raised to the fourth power one over three raised to the power of negative four −3^4 3^3

Narad:

\[(\frac{ 1 }{ 3 })^{4}=\frac{ 1 }{ 3 ^{4}}\] \[(\frac{ 1 }{ 3})^{-4}=\frac{ 1 }{ 3^{-4} }\]

Joe348:

Simplifying (3^4)-1 \[3 \times 3 \times 3 \times 3\] =\[81\] \[81-1\] = \[80\]

Joe348:

\[-3^4\] = \[-3 \times 3 \times 3 \times 3\] \[=-81\] \[3^3= 3 \times 3 \times 3 = 27\]

surjithayer:

\[(3^4)^{-1}=3^{(4\times -1)}=3^{-4}=\frac{ 1 }{ 3^4}\] a

jhonyy9:

@surjithayer wrote:
\[(3^4)^{-1}=3^{(4\times -1)}=3^{-4}=\frac{ 1 }{ 3^4}\] a
-1 is exponent sure ?

jhonyy9:

there is \[\left( 3^{4} \right)^{-1} = ? \] or \[3^{4} -1 = ?\]

Joe348:

@jhonyy9 wrote:
there is \[\left( 3^{4} \right)^{-1} = ? \] or \[3^{4} -1 = ?\]
\[(3^4)^-1 = 1/81\] \[3^4-1=80\] The second choice is the only correct answer.

Joe348:

@surjithayer wrote:
\[(3^4)^{-1}=3^{(4\times -1)}=3^{-4}=\frac{ 1 }{ 3^4}\] a
\[1/3^4=1/81\] Which is right for that problem, but not for this question.

jhonyy9:

1 over 3 mean negativ exponent of 3 and again negativ exponent of 3 not possibile

jhonyy9:

@joe348 wrote:
@jhonyy9 wrote:
there is \[\left( 3^{4} \right)^{-1} = ? \] or \[3^{4} -1 = ?\]
\[(3^4)^-1 = 1/81\] \[3^4-1=80\] The second choice is the only correct answer.
do you think on this one over three raised to the fourth power ?

Joe348:

No it would still be 1/81

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