How do you simplify ${(25 b^{6})}^{- 1.5}$ $(25b^6)^-1.5$ ?

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How do you simplify ${(25 b^{6})}^{- 1.5}$ $(25b^6)^-1.5$ ?

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Answer 1 · 2017-04-29T20:12:45

See the entire solution process below:

Explanation:

First, use these rules of exponents to remove the outer exponent:

$a = a^{1}$ $a = a^color(red)(1)$ and ${(x^{a})}^{b} = x^{a \times b}$ $(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))$

${(25 b^{6})}^{- 1.5} = {(25^{1} b^{6})}^{- 1.5} = 25^{1 \times - 1.5} b^{6 \times - 1.5} =$ $(25b^6)^-1.5 = (25^color(red)(1)b^color(red)(6))^color(blue)(-1.5) = 25^(color(red)(1) xx color(blue)(-1.5))b^(color(red)(6) xx color(blue)(-1.5)) =$

$25^{- 1.5} b^{- 6.5}$ $25^-1.5b^-6.5$

We can now use this rule of exponents to eliminate the negative exponents:

$25^{- 1.5} b^{- 6.5} = \frac{1}{25^{- - 1.5} b^{- - 6.5}} = \frac{1}{25^{1.5} b^{6.5}}$ $25^color(red)(-1.5)b^color(red)(-6.5) = 1/(25^color(red)(- -1.5)b^color(red)(- -6.5)) = 1/(25^1.5b^6.5)$

We can change the fractions to fractions as follows:

$\frac{1}{25^{1.5} b^{6.5}} = \frac{1}{25^{\frac{3}{2}} b^{\frac{13}{2}}}$ $1/(25^1.5b^6.5) = 1/(25^(3/2)b^(13/2))$

We can rewrite this expression as:

$\frac{1}{25^{\frac{1}{2} \times 3} b^{6.5}}$ $1/(25^(1/2 xx 3)b^6.5)$

We can rewrite this as:

$\frac{1}{{(25^{\frac{1}{2}})}^{3} b^{6.5}} = \frac{1}{5^{3} b^{6.5}} = \frac{1}{125 b^{6.5}}$ $1/((25^(1/2))^3b^6.5) = 1/(5^3b^6.5) = 1/(125b^6.5)$

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How do you simplify ${(25 b^{6})}^{- 1.5}$ $(25b^6)^-1.5$ ?

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How do you simplify ${(25 b^{6})}^{- 1.5}$ $(25b^6)^-1.5$ ?