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Please find the original thread, and answer here: https://web2.0calc.com/questions/manipulating-exponents

 

Melody the question I wrote was wrong. It was actually:

 

Write \(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\)as a power of 2. 

 Jan 18, 2022
 #1
avatar+117849 
0

Thanks Mathy.

Now it comes out perfectly well.

If you still need help with it please put this correct version, along with the explantatory note,  on the original thread.   

I can help you over there is you still want me too,

 Jan 18, 2022
 #2
avatar+193 
-1

 

Yes please do, much apperciated. 

MathyGoo13  Jan 19, 2022
 #3
avatar+117849 
+1

 

\(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\\ =\frac{4}{27}\left( \left[ 2^{\frac{3+m}{3}} \right] + \left[ 2^{\frac{m}{3}} \right] \right)^3\\ =\frac{4}{27}\left(\left[ 2^{\frac{3+m}{3}} \right]^3 + 3*\left[ 2^{\frac{3+m}{3}} \right]^2 \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] ^2 + \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+2m}{3}} \right] \left[ 2^{\frac{m}{3}} \right] + 3*\left[ 2^{\frac{3+m}{3}} \right] \left[ 2^{\frac{2m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{\frac{6+3m}{3}} \right] + 3*\left[ 2^{\frac{3+3m}{3}} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4}{27}\left(\left[ 2^{3+m} \right] + 3*\left[ 2^{2+m} \right] + 3*\left[ 2^{1+m} \right] + \left[ 2^{m} \right] \right)\\ =\frac{4*2^m}{27}\left(\left[ 2^{3} \right] + 3*\left[ 2^{2} \right] + 3*\left[ 2 \right] + \left[ 1 \right] \right)\\ =\frac{4*2^m}{27}\left(8+ 12 + 6 + 1\right)\\ =\frac{4*2^m}{27}\left(27\right)\\ =2^{(m+2)} \)

 

 

 

 

 

LaTex

\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{m}{3}}\right)^3\\

=\frac{4}{27}\left(     \left[  2^{\frac{3+m}{3}} \right]    +    \left[ 2^{\frac{m}{3}} \right]    \right)^3\\
=\frac{4}{27}\left(\left[  2^{\frac{3+m}{3}} \right]^3 
+ 3*\left[  2^{\frac{3+m}{3}} \right]^2   \left[ 2^{\frac{m}{3}} \right]     
+ 3*\left[  2^{\frac{3+m}{3}} \right]   \left[ 2^{\frac{m}{3}} \right]  ^2  +  \left[ 2^{\frac{m}{3}} \right] ^3 \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{\frac{6+2m}{3}} \right]   \left[ 2^{\frac{m}{3}} \right]     
+ 3*\left[  2^{\frac{3+m}{3}} \right]   \left[ 2^{\frac{2m}{3}} \right]   +  \left[ 2^{m} \right]  \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{\frac{6+3m}{3}} \right]      
+ 3*\left[  2^{\frac{3+3m}{3}} \right]     +  \left[ 2^{m} \right]  \right)\\

=\frac{4}{27}\left(\left[  2^{3+m} \right] 
+ 3*\left[  2^{2+m} \right]      
+ 3*\left[  2^{1+m} \right]     +  \left[ 2^{m} \right]  \right)\\

=\frac{4*2^m}{27}\left(\left[  2^{3} \right] 
+ 3*\left[  2^{2} \right]      
+ 3*\left[  2 \right]     +  \left[ 1 \right]  \right)\\

=\frac{4*2^m}{27}\left(8+  12 + 6    + 1\right)\\
=\frac{4*2^m}{27}\left(27\right)\\
=2^{(m+2)}

 Jan 19, 2022
 #4
avatar+117849 
0

Normally I would not have answered this on athis new thread, but the original unintentioned question was a completely different question.

 Jan 19, 2022

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