If $n$ is a hopeful integer and also $x$ is any type of real number, climate $x^n$ synchronizes to repeated multiplication \begingather* x^n = \underbracex \times x \times \cdots \times x_n \text times.\endgather*We can call this “$x$ elevated to the strength of $n$,” “$x$ to the power of $n$,” or merely “$x$ come the $n$.” Here, $x$ is the base and $n$ is the exponent or the power.

You are watching: 3 to the power of -1

From this definition, we can deduce some simple rules that exponentiation have to follow and some hand special instances that monitor from the rules. In the process, we"ll specify exponentials $x^a$ for exponents $a$ that aren"t hopeful integers.

The rules and special instances are summary in the adhering to table. Below, we provide details for each one.

Rule or special caseFormulaExample
Product$x^ax^b = x^a+b$$2^22^3 = 2^5=32 Quotient\displaystyle \fracx^ax^b = x^a-b$$\displaystyle \frac2^32^2 = 2^1 =2$
Power that power$(x^a)^b = x^ab$$(2^3)^2 = 2^6=64 Power of a product(xy)^a = x^ay^a$$36=6^2=(2\cdotbadbreak 3)^2 = 2^2\cdotbadbreak 3^2=4 \cdotbadbreak 9=36$
Power of one$x^1=x$$2^1=2 Power the zerox^0=1$$2^0=1$
Power of negative one$\displaystyle x^-1=\frac1x$$\displaystyle 2^-1=\frac12 Change authorize of exponents\displaystyle x^-a = \frac1x^a$$\displaystyle 2^-3 = \frac12^3 = \frac18$
Fractional exponentsx^m/n = \sqrtx^m = (\sqrtx)^m4^3/2 = (\sqrt4)^3=2^3=8 The rulesProduct of exponentials with very same base If we take the product of 2 exponentials with the exact same base, we simply add the exponents:\begingatherx^ax^b = x^a+b. \labelproduct\endgather To watch this rule, we just increase out what the index number mean. Let"s begin out through a pair simple examples.\beginalign* 3^4 3^2 &= (3 \times 3 \times 3 \times 3) \times (3 \times 3)\\ &= 3 \times 3 \times 3 \times 3 \times 3 \times 3\\ &= 3^6\endalign*\beginalign* y^2 y^3 &= (y \times y) \times (y \times y \times y)\\ &= y \times y \times y \times y \times y\\ & = y^5\endalign* The general situation works the very same way. We simply need to keep track that the variety of factors us have.\beginalign*x^ax^b &= \underbracex \times \cdots \times x_a \text times \times \underbracex \times \cdots \times x_b \text times\\<0.2cm> &= \underbracex \times \cdots \times x_a+b \text times\\<0.2cm> &=x^a+b\endalign* Quotient of exponentials with same base If us take the quotient of two exponentials with the exact same base, we simply subtract the exponents:\begingather\fracx^ax^b = x^a-b \labelquotient\endgather This preeminence results native canceling usual factors in the numerator and also denominator. Because that example:\beginalign* \fracy^5y^3 &= \fracy \times y \times y \times y \times yy \times y \times y\\ &= \frac(y \times y) \times \cancel(y \times y \times y)\cancely \times y \times y\\ &= y \times y = y^2.\endalign*To display this in general, we look at two various cases. If us imagine that a > b, climate this dominance follows native canceling the common b components of x that occur in both the numerator and denominator. We space left with just b-a determinants of x in the numerator.\beginalign*\fracx^ax^b &= \frac\quad \overbracex \times \cdots \times x^a \text times\quad\underbracex \times \cdots \times x_b \text times\\<0.2cm> &= \frac\quad \overbracex \times \cdots \times x^a-b \text times\times\overbrace\cancelx \times \cdots \times x^b \text times\quad\underbrace\cancelx \times \cdots \times x_b \text times\\<0.2cm> &= \underbracex \times \cdots \times x_a-b \text times\\<0.2cm> &=x^a-b\endalign* If a a, the exponent a-b is a negative number. Since formula \eqrefquotient is the exact same no matter the relationship between a and also b, we don"t must worry about it and also can simply subtract the exponents. Power that a power We can raise exponential to one more power, or take it a power of a power. The result is a single exponential whereby the strength is the product that the initial exponents:\begingather(x^a)^b = x^ab. \labelpower_power\endgather We have the right to see this an outcome by composing it as a product where the x^a is repeated b times:\begingather*(x^a)^b = \underbracex^a \times x^a \times \cdots \times x^a_b\text times.\endgather*Next we apply rule \eqrefproduct because that the product of exponentials through the exact same base. We use this dominion b times to finish that\beginalign*(x^a)^b &= \underbracex^a \times x^a \times \cdots \times x^a_b\text times\\<0.2cm> &= x^\overbracea + a + \cdots + a^b\text times\\<0.2cm> &= x^ab.\endalign*In the last step, we had actually to remember that multiplication deserve to be identified as repetitive addition. Power of a product If us take the power of a product, we have the right to distribute the exponent over the different factors:\begingather (xy)^a = x^ay^a. \labelpower_product\endgather We can present this rule in the same way as we present that you deserve to distribute multiplication end addition. One way to display this distributive law for multiplication is is come remember the multiplication is identified as repetitive addition:\beginalign* (x+y)a &= \underbrace(x + y) + (x+y) + \cdots + (x+y)_a\text times\\<0.2cm> &= \underbracex + x + \cdots + x_a\text times+\underbracey+ y + \cdots + y_a\text times\\<0.2cm>\\ &= xa +ya.\endalign*In the same way, us can show the distributive law for exponentiation:\beginalign* (xy)^a &= \underbrace(xy) \times (xy) \times \cdots \times (xy)_a\text times\\<0.2cm> &= \underbracex \times x \times \cdots \times x_a\text times\times\underbracey \times y \times \cdots \times y_a\text times\\<0.2cm>\\ &= x^a y^a.\endalign* This rule additionally works because that quotients\begingather* \left(\fracxy\right)^a = \fracx^ay^a,\endgather*but the does NOT job-related for sums. Because that example, \beginalign* (3+5)^2 = 8^2 = 64,\endalign*but this is no equal to \beginalign* 3^2+5^2 = 9 + 25 =34.\endalign* Special cases The complying with are special instances that monitor from the rules. The power of one The simplest special case is the raising any kind of number come the strength of 1 doesn"t do anything:\begingatherx^1=x.\labelpower_one\endgather The strength of zero As lengthy as x isn"t zero, elevating it come the strength of zero have to be 1:x^0=1.$$We can see this, because that example, native the quotient rule, as$$1 = \fracx^ax^a = x^a-a=x^0.$$The expression 0^0 is indeterminate. You have the right to see that it should be indeterminate, due to the fact that you deserve to come increase with great reasons because that it to it is in two various values. First, native above, if x \ne 0, then x^0=1, no matter how small x is. If we just let x go all the means to zero (take the limit as x goes to zero), then it seems that 0^0 need to be 1. On the other hand, 0^a=0 as lengthy as a \ne 0. Repetitive multiplication of 0 still offers zero, and we have the right to use the above rules to show 0^a tho is zero, no matter how little a is, as long as that is nonzero. If just let a go all the method to zero (take the limit as a goes to zero), then it seems choose 0^0 need to be 0. See more: Who Is The Only Fictional Character With Own Zip Code ? Geek Trivia: Zip Code Of Honor In various other words, if we start with x^a for non-zero x and non-zero a, we"ll obtain a different answer because that 0^0 relying on whether we let x walk to zero an initial or a go to zero first. There really is no means for deciding top top a value for 0^0, so us are forced to leaving it indeterminate. Friend can check out this applet come visualize this argument. The power of an unfavorable one Negative one is a one-of-a-kind value for an exponent, because taking a number to the power of an adverse one offers its reciprocal:$$x^-1 = \frac1x.$$The an altering sign of exponent In a similar vein, transforming the sign of a exponent gives the reciprocal, so$$x^-a = \frac1x^a.$Fractional exponents The strength of power ascendancy \eqrefpower_power permits us to define fractional exponents. For example, dominion \eqrefpower_power tells us that\begingather* 9^1/2=(3^2)^1/2 = 3^2 \cdot 1/2 = 3^1 = 3.\endgather*Taking a number come the power of$\frac12$undoes taking a number to the strength of 2 (or squaring it). In various other words, taking a number come the strength of$\frac12$is the exact same thing as taking a square root:\begingather* x^1/2 = \sqrtx.\endgather* Since$(x^n)^1/n = x^1 = x$, we can generalize the an outcome so that taking a number come the power of$1/n$is the same thing as acquisition the$n$th root:\begingather x^1/n = \sqrtx.\endgather If$a$is any kind of rational number, then it deserve to be composed as$a=m/n$. Us can define taking a number to the$a$th strength as taking that number to the$m$th power and also the$n$th root. We"ll assume the base$x$is non-negative for this reason that we don"t have to worry around doing things like taking the square source of a negative number. Then, the order doesn"t issue and\begingather* x^m/n = \sqrtx^m = (\sqrtx)^m.\endgather*If$a$is an irrational number, favor$a=\pi$, climate this process doesn"t precisely work. But, due to the fact that you can discover a rational number together close together you desire to any irrational number, you deserve to approximate$x^a$as well as you like. (To be precise, you can define$x^a$in terms of a border of$x^b$, wherein$b$space rational numbers approaching$a\$.)