Simplify the expression

chijioke

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Simplify the expression

Solution

\(\displaystyle \frac{10^\frac{3n}{2} × 15^\frac{n}{2}×6^\frac{n}{6}}{45^\frac{n}{3} × 20^\frac{2n}{3}}\)

[imath]\begin{array}{cc} \frac{10^\frac{3n}{2} × 15^\frac{n}{2}×6^\frac{n}{6}}{45^\frac{n}{3} × 20^\frac{2n}{3}} = \\[10pt] \newline \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 45 \times 20 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}} \end{array} [/imath]

Can it be simplified further?
 

Steven G

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Dec 30, 2014
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Some of the bases in the numerator and denominator have some common factors. What can you do with them?

You definitely should have put 400 instead of 20^2 but maybe some factors will cancel out from the 20 and ???.

In the numerator, you should have raised the base to n/6. That way you don't have things like 61/3 in the base. However, you should reduce common factors first. You should 1st factor all the factors in the base.
 

chijioke

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Some of the bases in the numerator and denominator have some common factors. What can you do with them?
I have already factored out the common factors I can see : \(\displaystyle \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 45 \times 20 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}}\)
You definitely should have put 400 instead of 20^2 but maybe some factors will cancel out from the 20 and ???.
You mean
\(\displaystyle \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n }{ 2 }}}{ \left( 45 \times 400 \right) ^ {\frac{ n }{ 3 }}}\)
I still can't find the factors that will cancel out.

In the numerator, you should have raised the base to n/6. That way you don't have things like 61/3 in the base. However, you should reduce common factors first. You should 1st factor all the factors in the base.
But \(\displaystyle \frac{n}{6}\) is not common to the terms in the numerator.
 
D

Deleted member 4993

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I have already factored out the common factors I can see : \(\displaystyle \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 45 \times 20 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}}\)

You mean
\(\displaystyle \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n }{ 2 }}}{ \left( 45 \times 400 \right) ^ {\frac{ n }{ 3 }}}\)
I still can't find the factors that will cancel out.


But \(\displaystyle \frac{n}{6}\) is not common to the terms in the numerator.
Factorize all the numbers to their "prime" factors - e.g. 45 = 32 * 5 and 15 = 3 * 5 and......
 

chijioke

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Factorize all the numbers to their "prime" factors - e.g. 45 = 32 * 5 and 15 = 3 * 5 and......

[imath]\begin{array}{c} \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 45 \times 20 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}} \\[8pt] =\frac{ \left( \left( 2 \times 5 \right) ^ { 3 } \times 3 \times 5 \times \left( 2 \times 3 \right) ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 \times \left( 2 ^ { 2 } \times 5 \right) ^ { 2 } \right) ^ {\frac{ n}{ 3 }}} \\[8pt] =\frac{ \left( 2 ^ { 3 } \times 5 ^ { 3 } \times 3 \times 5 \times 2 ^ {\frac{ 1 }{ 3 }} \times 3 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 3 ^ { 2 } \times 5 \times 2 ^ { 4 } \times 5 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}} \\[8pt] =\frac{ \left( 2 ^ { 3 } \times 2 ^ {\frac{ 1 }{ 3 }} \times 3 \times 3 ^ {\frac{ 1 }{ 3 }} \times 5 ^ { 3 } \times 5 \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 \times 5 ^ { 2 } \times 2 ^ { 4 } \right) ^ {\frac{ n}{ 3 }}} \\[8pt] =\frac{ \left( 2 ^ {\frac{ 10 }{ 3 }} \times 3 ^ {\frac{ 4 }{ 3 }} \times 5 ^ { 4 } \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 ^ { 3 } \times 2 ^ { 4 } \right) ^ {\frac{ n }{ 3 }}} \end{array}[/imath]
I can't go further. Is further simplification possible. I think need help here if am to go further .
 

JeffM

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Yes, you can go further, but it might be easier to do it this way

[math] 10^{3n/2} * 15^{n/2} * 6^{n/6} = 10^{9n/6} * 15^{3n/6} * 6^{n/6} = (10^9 * 15*3 * 6)^{n/6} =\\ (5^9 * 2^9 * 5^3 * 3^3 * 3 * 2)^{n/6} = (5^{12} * 3^4 * 2^{10})^{n/6}.\\ 45^{n/3} * 20^{2n/3} = 45^{2n/6} * 20^{4n/6} = ( 45^2 * 20^4)^{n/6} = (3^4 * 5^2 * 5^4 * 2^8)^{n/6} = (3^4 * 5^6 * 2^8)^{n/6}.\\ \therefore \ \dfrac{10^{3n/2} * 15^{n/2} * 6^{n/6}}{45^{n/3} * 20^{2n/3}} = \left ( \dfrac{5^{12} * 3^4 * 2^{10}}{3^4 * 5^6 * 2^8}\right )^{n/6} =\\ (5^6 * 2^2)^{n/6} = (5 \sqrt[3]{2} )^n. [/math]
 

Steven G

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[imath]\begin{array}{c} \frac{ \left( 10 ^ { 3 } \times 15 \times 6 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 45 \times 20 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}} \\[8pt] =\frac{ \left( \left( 2 \times 5 \right) ^ { 3 } \times 3 \times 5 \times \left( 2 \times 3 \right) ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 \times \left( 2 ^ { 2 } \times 5 \right) ^ { 2 } \right) ^ {\frac{ n}{ 3 }}} \\[8pt] =\frac{ \left( 2 ^ { 3 } \times 5 ^ { 3 } \times 3 \times 5 \times 2 ^ {\frac{ 1 }{ 3 }} \times 3 ^ {\frac{ 1 }{ 3 }} \right) ^ {\frac{ n}{ 2 }}}{ \left( 3 ^ { 2 } \times 5 \times 2 ^ { 4 } \times 5 ^ { 2 } \right) ^ {\frac{ n }{ 3 }}} \\[8pt] =\frac{ \left( 2 ^ { 3 } \times 2 ^ {\frac{ 1 }{ 3 }} \times 3 \times 3 ^ {\frac{ 1 }{ 3 }} \times 5 ^ { 3 } \times 5 \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 \times 5 ^ { 2 } \times 2 ^ { 4 } \right) ^ {\frac{ n}{ 3 }}} \\[8pt] =\frac{ \left( 2 ^ {\frac{ 10 }{ 3 }} \times 3 ^ {\frac{ 4 }{ 3 }} \times 5 ^ { 4 } \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 ^ { 3 } \times 2 ^ { 4 } \right) ^ {\frac{ n }{ 3 }}} \end{array}[/imath]
I can't go further. Is further simplification possible. I think need help here if am to go further .
Now multiply each exponent by the outside power and then simplify the common factors from top to bottom
 

chijioke

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Now multiply each exponent by the outside power and then simplify the common factors from top to bottom
Continuing from where I stopped.
[math] =\frac{ \left( 2 ^ {\frac{ 10 }{ 3 }} \times 3 ^ {\frac{ 4 }{ 3 }} \times 5 ^ { 4 } \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 ^ { 3 } \times 2 ^ { 4 } \right) ^ {\frac{ n }{ 3 }}} \\[4pt] = \frac{ 2 ^ { \left( \frac{ 10 }{ 3 } \times \frac{ n }{ 2 } \right) } \times 3 ^ { \left( \frac{ 4 }{ 3 } \times \frac{ n }{ 2 } \right) } \times 5 ^ { \left( 4 \times \frac{ n }{ 2 } \right) }}{ 3 ^ { \left( 2 \times \frac{ n}{ 3 } \right) } \times 5 ^ { \left( 3 \times \frac{ n}{ 3 } \right) } \times 2 ^ { \left( 4 \times \frac{ n }{ 3 } \right) }} \\[4pt] = \frac{ 2 ^ {\frac{ 10 \times n}{ 6 }} \times 3 ^ {\frac{ 4 \times n }{ 6 }} \times 5 ^ {\frac{ 4 \times n }{ 2 }}}{ 3 ^ {\frac{ 2 \times n }{ 3 }} \times 5 ^ {\frac{ 3 \times n }{ 3 }} \times 2 ^ {\frac{ 4 \times n}{ 3 }}} \\[4pt] = \frac{ 2 ^ {\frac{ 10 \times n }{ 6 }}}{ 2 ^ {\frac{ 4 \times n }{ 3 }}} \times \frac{ 3 ^ {\frac{ 4 \times n }{ 6 }}}{ 3 ^ {\frac{ 2 \times n }{ 3 }}} \times \frac{ 5 ^ {\frac{ 4 \times n }{ 2 }}}{ 5 ^ {\frac{ 3 \times n }{ 3 }}} \\[4pt] = 2 ^ { \left( \frac{ 10 \times n }{ 6 } - \frac{ 4 \times n}{ 3 } \right) } \times 3 ^ { \left( \frac{ 4 \times n }{ 6 } - \frac{ 2 \times n }{ 3 } \right) } \times 5 ^ { \left( \frac{ 4n}{ 2 } - \frac{ 3 \times n}{ 3 } \right) } \\[3pt] = 2 ^ {\frac{n }{ 3 }} \times 3 ^ { 0 } \times 5 ^ { n } \\[3pt] = 2 ^ {\frac{ n}{ 3 }} \times 1 \times 5 ^ { n} \\[3pt] = 2 ^ {\frac{ n}{ 3 }} \times 5 ^ { n } \\[3pt] = \left(\sqrt[ 3 ]{ 2 }\right) ^ { n } \times 5 ^ { n} \\[3pt] = \left( 5 \sqrt[ 3 ]{ 2 } \right) ^ { n } [/math]
 
D

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Continuing from where I stopped.
[math] =\frac{ \left( 2 ^ {\frac{ 10 }{ 3 }} \times 3 ^ {\frac{ 4 }{ 3 }} \times 5 ^ { 4 } \right) ^ {\frac{ n }{ 2 }}}{ \left( 3 ^ { 2 } \times 5 ^ { 3 } \times 2 ^ { 4 } \right) ^ {\frac{ n }{ 3 }}} \\[4pt] = \frac{ 2 ^ { \left( \frac{ 10 }{ 3 } \times \frac{ n }{ 2 } \right) } \times 3 ^ { \left( \frac{ 4 }{ 3 } \times \frac{ n }{ 2 } \right) } \times 5 ^ { \left( 4 \times \frac{ n }{ 2 } \right) }}{ 3 ^ { \left( 2 \times \frac{ n}{ 3 } \right) } \times 5 ^ { \left( 3 \times \frac{ n}{ 3 } \right) } \times 2 ^ { \left( 4 \times \frac{ n }{ 3 } \right) }} \\[4pt] = \frac{ 2 ^ {\frac{ 10 \times n}{ 6 }} \times 3 ^ {\frac{ 4 \times n }{ 6 }} \times 5 ^ {\frac{ 4 \times n }{ 2 }}}{ 3 ^ {\frac{ 2 \times n }{ 3 }} \times 5 ^ {\frac{ 3 \times n }{ 3 }} \times 2 ^ {\frac{ 4 \times n}{ 3 }}} \\[4pt] = \frac{ 2 ^ {\frac{ 10 \times n }{ 6 }}}{ 2 ^ {\frac{ 4 \times n }{ 3 }}} \times \frac{ 3 ^ {\frac{ 4 \times n }{ 6 }}}{ 3 ^ {\frac{ 2 \times n }{ 3 }}} \times \frac{ 5 ^ {\frac{ 4 \times n }{ 2 }}}{ 5 ^ {\frac{ 3 \times n }{ 3 }}} \\[4pt] = 2 ^ { \left( \frac{ 10 \times n }{ 6 } - \frac{ 4 \times n}{ 3 } \right) } \times 3 ^ { \left( \frac{ 4 \times n }{ 6 } - \frac{ 2 \times n }{ 3 } \right) } \times 5 ^ { \left( \frac{ 4n}{ 2 } - \frac{ 3 \times n}{ 3 } \right) } \\[3pt] = 2 ^ {\frac{n }{ 3 }} \times 3 ^ { 0 } \times 5 ^ { n } \\[3pt] = 2 ^ {\frac{ n}{ 3 }} \times 1 \times 5 ^ { n} \\[3pt] = 2 ^ {\frac{ n}{ 3 }} \times 5 ^ { n } \\[3pt] = \left(\sqrt[ 3 ]{ 2 }\right) ^ { n } \times 5 ^ { n} \\[3pt] = \left( 5 \sqrt[ 3 ]{ 2 } \right) ^ { n } [/math]
Looks goooood to me.......
 

chijioke

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Tha
Yes, you can go further, but it might be easier to do it this way

[math] 10^{3n/2} * 15^{n/2} * 6^{n/6} = 10^{9n/6} * 15^{3n/6} * 6^{n/6} = (10^9 * 15*3 * 6)^{n/6} =\\ (5^9 * 2^9 * 5^3 * 3^3 * 3 * 2)^{n/6} = (5^{12} * 3^4 * 2^{10})^{n/6}.\\ 45^{n/3} * 20^{2n/3} = 45^{2n/6} * 20^{4n/6} = ( 45^2 * 20^4)^{n/6} = (3^4 * 5^2 * 5^4 * 2^8)^{n/6} = (3^4 * 5^6 * 2^8)^{n/6}.\\ \therefore \ \dfrac{10^{3n/2} * 15^{n/2} * 6^{n/6}}{45^{n/3} * 20^{2n/3}} = \left ( \dfrac{5^{12} * 3^4 * 2^{10}}{3^4 * 5^6 * 2^8}\right )^{n/6} =\\ (5^6 * 2^2)^{n/6} = (5 \sqrt[3]{2} )^n. [/math]
What did do or think to know that
[math]\frac{n}{6}[/math] can be used to factor the expression?
 

chijioke

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What is the GCF of (n/2 , n/3)?
That is interesting. I never knew that GCF of fractions can be found as well. I used to think that is only that of monomials that can be found.
I just googled how to find the GCF of fractions, the formula is
[math]\frac{\text{gcf of numerator}}{\text{lcm of denominator}}[/math]I just applied the formula and discovered that GCF of
[math]\frac{n}{2}~\text{and}~\frac{n}{3} = \frac{n}{6}[/math]Thank you.
 

JeffM

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That is interesting. I never knew that GCF of fractions can be found as well. I used to think that is only that of monomials that can be found.
I just googled how to find the GCF of fractions, the formula is
[math]\frac{\text{gcf of numerator}}{\text{lcm of denominator}}[/math]I just applied the formula and discovered that GCF of
[math]\frac{n}{2}~\text{and}~\frac{n}{3} = \frac{n}{6}[/math]Thank you.
And now you may understand why I suggested that there might be a less arduous approach. Sometime the order in which you simplify is computationally helpful.
 
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