What is e^{-ln((v^4(4-u)^4)/(u^4(4-v)^4))} ?

The solution to e^{-ln((v^4(4-u)^4)/(u^4(4-v)^4))} is (u^4(-v+4)^4)/(v^4(-u+4)^4)

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simplify e^{-ln((v^4(4-u)^4)/(u^4(4-v)^4))}

simplify

e^{- l n (\frac{v ^{4} ( 4 - u ) ^{4}}{u ^{4} ( 4 - v ) ^{4}})}

\frac{u ^{4} ( - v + 4 ) ^{4}}{v ^{4} ( - u + 4 ) ^{4}}

Solution steps

e^{- l n (\frac{v ^{4} ( 4 - u ) ^{4}}{u ^{4} ( 4 - v ) ^{4}})}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{l n (\frac{v ^{4} ( 4 - u ) ^{4}}{u ^{4} ( 4 - v ) ^{4}})}}

e^{l n (\frac{v ^{4} ( 4 - u ) ^{4}}{u ^{4} ( 4 - v ) ^{4}})} = \frac{v ^{4} ( 4 - u ) ^{4}}{u ^{4} ( 4 - v ) ^{4}}

= \frac{1}{\frac{v ^{4} ( - u + 4 ) ^{4}}{u ^{4} ( - v + 4 ) ^{4}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{u ^{4} ( 4 - v ) ^{4}}{v ^{4} ( 4 - u ) ^{4}}

2.511 \cdot 1 0^{- 7}

3.12 \cdot 1 0^{- 6}

0.024 1^{- \frac{1}{2}} \cdot 1.3293400000000 \dots

(\frac{6.9}{3}) \cdot (\frac{1 0 ^{- 1}}{1 0 ^{3}})

What is e^{-ln((v^4(4-u)^4)/(u^4(4-v)^4))} ?
The solution to e^{-ln((v^4(4-u)^4)/(u^4(4-v)^4))} is (u^4(-v+4)^4)/(v^4(-u+4)^4)