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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
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\[(\frac{ 1 }{ 3 })^{4}=\frac{ 1 }{ 3 ^{4}}\] \[(\frac{ 1 }{ 3})^{-4}=\frac{ 1 }{ 3^{-4} }\]
Simplifying (3^4)-1 \[3 \times 3 \times 3 \times 3\] =\[81\] \[81-1\] = \[80\]
\[-3^4\] = \[-3 \times 3 \times 3 \times 3\] \[=-81\] \[3^3= 3 \times 3 \times 3 = 27\]
\[(3^4)^{-1}=3^{(4\times -1)}=3^{-4}=\frac{ 1 }{ 3^4}\] a
@surjithayer wrote:
\[(3^4)^{-1}=3^{(4\times -1)}=3^{-4}=\frac{ 1 }{ 3^4}\]
a
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there is \[\left( 3^{4} \right)^{-1} = ? \] or \[3^{4} -1 = ?\]
@jhonyy9 wrote:
there is
\[\left( 3^{4} \right)^{-1} = ? \]
or
\[3^{4} -1 = ?\]
@surjithayer wrote:
\[(3^4)^{-1}=3^{(4\times -1)}=3^{-4}=\frac{ 1 }{ 3^4}\]
a
1 over 3 mean negativ exponent of 3 and again negativ exponent of 3 not possibile
@joe348 wrote:
@jhonyy9 wrote:
there is
\[\left( 3^{4} \right)^{-1} = ? \]
or
\[3^{4} -1 = ?\]
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No it would still be 1/81
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