Simplify the expression

chijioke

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343
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?
 
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.
 
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.
 
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......
 
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 .
 
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]
 
[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
 
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]
 
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.......
 
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?
 
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.
 
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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