What value of k solves the equation?

Question

anonymous · Answer

attachment

whpalmer4 · Answer

$$k^{-3} = \frac{1}{27}$$property of negative exponents:
$$x^{-n} = \frac{1}{x^n}$$

whpalmer4 · Answer

$$\frac{1}{k^3} = \frac{1}{27}$$$$k=$$

whpalmer4 · Answer

hint:
$$\frac{1}{k^3} = \frac{1}{k*k*k}$$

anonymous · Answer

@PeachRings @acxbox22

whpalmer4 · Answer

You have all the information you need to solve the problem.

whpalmer4 · Answer

what is 5*5*5=

anonymous · Answer

125

whpalmer4 · Answer

That's too big, isn't it?  Let's a try a smaller number.  What is 2*2*2=

acxbox22 · Answer

$$k ^{3}=27$$
$$\sqrt[3]{k^3}=\sqrt[3]{27}$$
k=3

anonymous · Answer

thanks all of you!

acxbox22 · Answer

@whpalmer4 if you continue doing guess and check you might as well do 2.5 x 2.5 x 2.5, and 2.6x2.6x2.6 and so on in obtaing 27. might as well use a calculator.

whpalmer4 · Answer

No, continue on with integers until you've hit the answer or the pair of integers that straddle it.  That's how you build up a good number sense, not by using a calculator.

whpalmer4 · Answer

And really, no one who has memorized the multiplication table up to 10x10 should have any trouble figuring out that 3x3x3 = 27!

acxbox22 · Answer

exactly..you either know the cubes or you dont
this was a simple question but would you go on with your method for bigger numbers
what if it was k^3=789 
i am pretty sure any wise person would use a calculator and not list ideas until you hit 9.24!

whpalmer4 · Answer

@acxbox22 This is a beginner problem, designed to teach them to use properties of exponents.  Once they have a handle on that, problems with answers that aren't so convenient become more appropriate.

You don't have to actually know the cubes to try multiplying 2*2*2, 3*3*3, etc. just as you don't have to know the squares to find the square root of 121 by guess and check.  Way too many students on OS appear to be completely helpless without some sort of computing assistance.  They need the practice that finding the answers like this gives them.

As for your problem, here's my calculator-free solution.
$9^3 = 9*9*9 = 81*9 =729$, so take $a=9$ and expand $$(a+b)^3 = a^3 + 3a^2b+3ab^2 + b^3$$
Disregard the terms with higher powers of $b$ and estimate $$789 = 9^3 + 3*9*9*b$$$$60=243b$$$$b\approx 0.25$$so $$\sqrt[3]{789}\approx 9+0.25=9.25$$for an error of about 1%.