MHB How can you simplify exponents involving fractions and roots?

  • Thread starter Thread starter CSmith1
  • Start date Start date
  • Tags Tags
    Algebra
Click For Summary
The discussion centers on simplifying exponents involving fractions and roots, specifically using the example of 32 raised to the power of 2/5. The calculations initially presented are critiqued for clarity and notation, emphasizing the importance of using proper formatting like LaTeX for better understanding. A simpler approach is suggested, noting that 32 can be expressed as 2^5, leading to the simplification of (2^5)^(2/5) to 2^2, which equals 4. The conversation highlights the strategy of applying roots before powers for easier calculations. Ultimately, the key takeaway is the importance of clarity in notation and the flexibility in the order of operations for simplification.
CSmith1
Messages
39
Reaction score
0
is this correct?
1.) 32 2/5(32 1/5) ^2=(5 SQUARE ROOT 32)2

=5(SQUAREROOT 2X2X2X2X2)2
=(5 SQUARE ROOT )2

=5 SQUARE ROOT 2 X 5 SQUARE ROOT 2
=25 (2 SQUARE ROOT 2)
=50 SQUARE ROOT 2
 
Mathematics news on Phys.org
CSmith said:
is this correct?
1.) $32^{2/5}$$(32^{1/5}) ^2$=(5 SQUARE ROOT 32)2
$(\sqrt[5]{32})^2$

=5(SQUAREROOT 2X2X2X2X2)2
$= (\sqrt[5]{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2})^2$

Notation aside you're correct up to here

=(5 SQUARE ROOT )2

=5 SQUARE ROOT 2 X 5 SQUARE ROOT 2
=25 (2 SQUARE ROOT 2)
=50 SQUARE ROOT 2

You've lost me here though. You have $(\sqrt[5]{2^5})^2$ which is the same as $((2^5)^{1/5})^2$ which cancels down to $2^2$

A simpler way IMO is to note that $32 = 2^5$ (as you did). That leaves you with $(2^5)^{2/5} = 2^{5 \cdot 2/5} = 2^2$Your notation is very tricky to follow. Please use Latex or "^" together with brackets next time to make it easer to follow.
 
I don't know how you were taught to approach these, but the short way to remember a fraction as an exponent is "power over root". The top number of the fraction is a power, so you should multiply the number together that many times. The bottom number is a root, so you need to find a number that multiplied together that many times gives you the original number. You can do it in either order, but you do one at a time. You have 32 to begin with so it's logical to take the root first because that will be a smaller number than 32. If you apply the power first then the result will be quite large and less easy to work with.

[math]32^{\frac{1}{5}}=2[/math] because $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 32$.

So now we've applied the root and got 2. Now we apply the power and get $2^2=4$ and we're done. Again you can choose the order in which you calculate this so take a second to consider both options and choose the one that has the easier numbers to work with.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...