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Use the properties of rational exponents to simplify the expression

$(3^\frac{1}3 \cdot 4^\frac{1}{4})^3$

I got $3^1 \cdot 4^\frac{3}{4}$

I just wasn't sure if this was the most simplified expression. It says to do it without a calculator so can someone confirm if I'm right?

QuantumPi
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  • You could get rid of the $1$ since $3^1=3$. You could also if you like reduce the $4$ some... noting that $4^{1/4} = 2^{1/2}$. If this makes it "simpler" is up to you. – JMoravitz Feb 04 '21 at 13:44
  • $4^{\frac{3}{4}} =4^{\frac{3}{2}\times \frac{1}{2}}= 2^{\frac{3}{2}} = 2\sqrt{2} $ – P. J. Feb 04 '21 at 13:44
  • A reminder to be cautious when trying to simplify exponents. This is valid in this case since everything we are working with are positive real numbers, however if you start throwing negative numbers or complex numbers into the mix, these properties you refer to may fail. – JMoravitz Feb 04 '21 at 13:46
  • I'm super confused now. – QuantumPi Feb 04 '21 at 13:47
  • About what? About my warning? About why $4^{1/4}=2^{1/2}$? About why $3^1=3$? – JMoravitz Feb 04 '21 at 13:49
  • no, I'm just generally worried about how I'm going to simplify this on the test :( – QuantumPi Feb 04 '21 at 13:50
  • You did fine in my opinion in your original attempt. "The most simplified expression" is incredibly subjective. Whether you consider $3\cdot 4^{3/4}$ or $6\sqrt{2}$ or $3\cdot 2^{3/2}$ or what to be "the most simplified" is personal preference. I wouldn't worry too much about it. – JMoravitz Feb 04 '21 at 13:52

2 Answers2

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$$(3^{1/3}\cdot 4^{1/4})^3$$

distribute the exponent using the rule that $(a\cdot b)^c = a^c\cdot b^c$ and the rule that $(a^b)^c = a^{(bc)}$ (which as alluded to are only guaranteed for when dealing with positive numbers). This leads us to:

$$(3^{3/3}\cdot 4^{3/4})$$

Simplify the fraction $3/3$ and recognize that this would leave it with an exponent of $1$ and exponents of $1$ can be omitted for brevity. This leads us to:

$$3\cdot 4^{3/4}$$

This is a perfectly fine place to stop if you like. If you prefer, you can continue on however. Recognizing $4$ as $2^2$, we could have continued as:

$$3\cdot (2^2)^{3/4}$$

$$3\cdot 2^{3/2}$$

and if you prefer writing things with surds

$$3\cdot 2\cdot \sqrt{2}$$

$$6\sqrt{2}$$

JMoravitz
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$$(3^\frac{1}3 \cdot 4^\frac{1}{4})^3$$ $$=(3^\frac{1}3 \cdot 4^\frac{1}{3}\cdot 4^\frac{-1}{12})^3$$ $$=(12^\frac{1}3\cdot 4^\frac{-1}{12})^3$$ $$=12(4^\frac{-1}{12})^3$$ $$=\frac{12}{(4^\frac{1}{12})^3}$$ $$=\frac{12}{4^\frac{1}{4}}$$ $$=\frac{12}{\sqrt{2}}$$ $$=6\sqrt{2}$$

So after simplifying the expression reduces to $$6\sqrt{2}$$

Smriti Sivakumar
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    @Klaus what are you referring to? Nothing seems amiss to me. It is many more steps to get to the end than I would have preferred, but the same final answer as is alluded to in comments above. – JMoravitz Feb 04 '21 at 13:47
  • @Klaus I believe this is perfectly correct.. – Smriti Sivakumar Feb 04 '21 at 13:47
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    @SmritiSivakumar I'm sorry, I misread the post. I deleted my (stupid) comment. – Klaus Feb 04 '21 at 13:51
  • I don't know how to do this on the test :sob: :Sob: – QuantumPi Feb 04 '21 at 13:51
  • @JMoravitz Ah yes, it is too many steps, just got carried away typing this :) Obviously you can just simplify from where it was left in the question – Smriti Sivakumar Feb 04 '21 at 13:52
  • @QuantumPi are you referring to questions of this model? because this is pretty simple if you are aware of basic arithmetic operations.. – Smriti Sivakumar Feb 04 '21 at 13:53
  • I can do the operations easily but I probably won't be big brain enough to think of breaking them into different pieces. – QuantumPi Feb 04 '21 at 13:54
  • @QuantumPi if this is with regard to some sort of examination, I think you should try and see if you can bring down the expression into some sort of combination of prime numbers. Ex. if the question is something like $(2)^{\frac{1}{2}}(14)^{\frac{1}{3}}$, split 14 into 2 and 7, group the two's, and try to simplify the exponents. Should work for typical high-school examination questions – Smriti Sivakumar Feb 04 '21 at 13:58
  • ohh ok. Thanks for da info – QuantumPi Feb 04 '21 at 13:59
  • @QuantumPi Whereas if you're talking in general, I think the comments answer your question – Smriti Sivakumar Feb 04 '21 at 13:59