# negative exponent problem

$$\sqrt{\frac{1}{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}$$

Does this equal =

\begin{align*} & \sqrt{3^0 + 3^1 + 3^2 + 3^3 + 3^4} \\ =&\sqrt{1 + 3 + 9 + 27 + 81} \\ =&\sqrt{121} \\ =&11. \end{align*}

The answer is apparently $\frac{9}{11}$ and I'm not sure what rule of negative exponents I got wrong.

The rule I'm using, incorrectly, is this:

$$\frac{1}{3^{-2}} = 3^2 = 9.$$

• You have the rule right, but you're forgetting that you can't just invert sums of fractions. You need to compute the sum in the denominator first before inverting. Do you see why? – Lost Dec 10 '13 at 5:37
• @Jwan622 please see my answer – dato datuashvili Dec 10 '13 at 5:40
• @Lost... I don't see why. – Jwan622 Dec 10 '13 at 7:12
• @ Lost. Is it because of Pemdas and the whole denominator is in parentheis... or at least implied parenthesis? Is that why? – Jwan622 Dec 10 '13 at 7:13
• @Jwan622 this is precisely why (at least this is one way of viewing your error) - order of basic operations in the real numbers...see my answer below – afedder Dec 10 '13 at 7:21

Note that $3^{-n} = 1/3^{n}$ for all $n \geq 0$, where $3^{0} = 1$, by convention. First, simplify the denominator utilizing the last fact: $$3^{0} + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} = 1 + 1/3 + 1/9 + 1/27 + 1/81 \\ = 81/81 + 27/81 + 9/81 + 3/81 + 1/81 \\ = 121/81$$ Thus, we have simplified the original expression to: $$\sqrt{1/(121/81)} = \sqrt{81/121} = 9/11$$ where the last equality comes from knowing perfect squares for natural numbers less than $20$ (recommended for any student of mathematics). Observe from the above that you had the correct "fact" all along, you simply need to recognize that each term of the denominator cannot be inverted separately. Explicitly, $$1/(3^{0} + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}) \neq 1/3^{0} + 1/3^{-1} + 1/3^{-2} + 1/3^{-3} + 1/3^{-4}$$

It might also be useful for you multiply the numerator and denominator by $\sqrt{3^{4}}$ as was mentioned in some of the other answers, but it is not too difficult to think about it without doing this in my opinion. Let me know if you have any further questions.

• How does one remember or know that facts like "each term of the denominator cannot be inverted separately?" – Jwan622 Dec 11 '13 at 5:08
• @Jwan622 It is simply a matter of order of operations (PEMDAS), as you claimed before. The numerator and the denominator of any fraction are contained in an "implied" set of parentheses. Thus, you should simplify these expressions first, in general, before proceeding to dealing with the fraction as a whole. – afedder Dec 11 '13 at 6:33
• @Jwan622 You can think of each basic operation ($+$, $-$, $\cdot$, $/$) as a function that maps a pair of real numbers to another real number. For example, multiplication is defined $$(a,b) \mapsto a \cdot b = \sum_{i=1}^{b} a = \sum_{i=1}^{a} b$$ for any $a,b \in \mathbb{R}$. If this is the framework we are considering, then using these operators is applying functions. But the simpler functions (i.e., $+$) are used to define the others, such as multiplication shown, so we can only apply in a certain order. From this sort of logic, we can see where order of operations originates. – afedder Dec 11 '13 at 6:59
• @Jwan622 note that the notation $\sum$ is the same as applying $+$ multiple times. To be more explicit, $$\sum_{i=1}^b a = a + a + a + ... + a$$ where $+$ was applied $b$ times. Similarly, $$\sum_{i=1}^a b = b + b + b + ... + b$$ where $+$ was applied $a$ times. – afedder Dec 11 '13 at 7:01
• @Jwan622 if the above discussion was a little too complicated, just remember order of operations in all circumstances when simplifying and you will be fine. In this case, the reason that I know you cannot invert each term of the denominator separately is that the denominator is its own separate expression and thus can only "interact" with the rest of the fraction as an entire "unit" (unless it is factored). – afedder Dec 11 '13 at 7:11

$$\sqrt{\frac1{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}=\sqrt{\frac{3^4}{3^4(3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4})}}$$

$$=\sqrt{\frac{3^4}{3^4+3^3+3^2+3^1+3^0}}=\sqrt{\frac{81}{121}}=\frac9{11}$$

• plus +1 for great answer – dato datuashvili Dec 10 '13 at 6:11

When you have a doubt with such expressions (I use here you specific problem), inside the square root, multiply both numerator and denominator by the reverse of the most negative power (here : 3^4) and perform the multiplications by (3^4) for each term of the denominator. Then, perform the additions and simplifications.

• please one for great answer +1 – dato datuashvili Dec 10 '13 at 6:12
• @datodatuashvili. Thanks ! You see, the (only) advantage of ages is to try to make things as simple as possible even if they are not the most fancy solutions. Cheers. – Claude Leibovici Dec 10 '13 at 6:29
• you are welcome,i think most advantage part of this is to help somebody,it does not matter how beautiful solution is – dato datuashvili Dec 10 '13 at 6:32

no you misses $1$ from of division,it is equal

$\sqrt{1/(1+1/3+1/9+1/27+1/81)}$

please find now least common multiple and simplify equation

just to simplify situation,least common multiple is $243$,we got it by $(81*27)/9$,

there $9$ is greatest common divisor,now let us solve it we would have

$(243+81+27+9+3)/243=363/243$

so we have $\sqrt {(1/(363/243)}$ or $\sqrt{243/363}$, we have common $3$,we we would have $\sqrt{81/121}$ and this is equal $9/11$

• $\text{lcm}(27,81) = 81 \neq 243$ – afedder Jun 7 '14 at 17:40
• aa,right yes thanks for point it – dato datuashvili Jun 7 '14 at 17:49
• No problem @datodatuashvili – afedder Jun 7 '14 at 17:50