Question

verify that [(729)-5/3]-1/2=(729) - 5/3*(-1/2)

Answer 1

Question -

$({(729)}^{ \frac{ - 5}{3} } ) ^{ \frac{ - 1}{2} } = {(729)}^{ \frac{ - 5}{3} \times \frac{ - 1}{2} }$


Answer -

LHS -

$({(729)}^{ \frac{ - 5}{3} })^{ \frac{ - 1}{2} }$

$we \: know \: ({a}^{m} ) ^{n} = {a}^{mn}$


$(729) ^{ \frac{5}{6} }$


$729 = {3}^{6}$

$({3}^{6}) ^{ \frac{5}{6} }$

${3}^{5} = 243$


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RHS -


$(729) ^{ \frac{ - 5}{3} \times \frac{ - 1}{2} }$

$(729) ^{ \frac{5}{6} }$

$729 = {3}^{6}$


$( {3}^{6} ) ^{ \frac{5}{6} }$

${3}^{5} = 243$


Hence :


Verified : LHS = RHS .

Answer 2

Answer:

Step-by-step explanation:

According to the power property of exponent

$(a^m)^n=a^{mn}$

The given equation is

$[(729)^{-\frac{5}{3}}]^{-\frac{1}{2}}=(729)^{-\frac{5}{3}\times (-\frac{1}{2})}$

We need to prove that LHS=RHS.

$LHS=[(729)^{-\frac{5}{3}}]^{\frac{1}{2}}$

$LHS=[(3^6)^{-\frac{5}{3}}]^{\frac{1}{2}}$

Using the power property of exponent, we get

$LHS=[(3^6\times (-\frac{5}{3}})]^{\frac{1}{2}}$

$LHS=(3^{-10})^{\frac{1}{2}}$

Using the power property of exponent, we get

$LHS=3^{(-10)\times \frac{1}{2}}$

$LHS=3^{5}$

$LHS=243$

Now, solve RHS of the given equation.

$RHS=(729)^{-\frac{5}{3}\times (-\frac{1}{2})}$

$RHS=(729)^{\frac{5}{6}}$

$RHS=(3^6)^{\frac{5}{6}}$

Using the power property of exponent, we get

$RHS=3^{6\times\frac{5}{6}}$

$RHS=3^5$

$RHS=243$

So, we get

$LHS=RHS$

Hence proved.