How do you solve powers of powers in advanced algebra

1 answer:

7 0

Ex: 2^2^2

Basically, the base is multiplied by the first power however large the number. Then the new base is multiplied by the second power to get an even larger number.
Say you have a question that asks for the power of the power of 2. Or 2^2^2.

You would do the equation 2^2 first, and end up with 4. But since you have another power, you would have the answer as 4^2. Which would then equal 16. But another equation that is similar is: 2^2^2 = 2^4
The two equations would still get the same answer, but would look entirely different.

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The length of the longer leg of a right triangle is 6 cm more than twice the length of the shorter leg. The length of the hypote

iragen [17]

Let x = the shorter leg
2x + 6 is the longer leg
2x + 9 is the hypotenuse

We now do the Pythagorean Theorem of $a^{2}+b^{2} = c^{2}$

$x^{2} + (2x+6)^{2} = (2x+9)^{2}$

Multiply out to get:

$x^{2} +4 x^{2} +24x+36=4 x^{2} +36x+81$

The $4 x^{2}$ will cancel out each other when you combine like terms to get all the terms on one side of the equal sign.

Combine like terms with the rest to end with $x^{2} -12x-45=0$

Factor this to get $(x-15)(x+3) = 0$ then set each factor equal to zero.

You end up with two solutions, x = 15 and x = - 3 HOWEVER distance cannot be measure in negatives so that answer will not work so x = 15. Now plug into our given information.

The short leg is 15 cm long.
The longer leg of 2(15) + 6 = 36 cm long
The hypotenuse of 2(15) + 9 = 39 cm long.

You can check your answers by plugging them into the Pythagorean Theorem again.

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3 years ago

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pashok25 [27]

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