$3^4 = 3 \cdot 3 \cdot 3 \cdot 3$

$3^2 = 3 \cdot 3$

$3^1 = 3$

But my problem is for below 1:


$3^{0.7} = \ ?$

Can you please explain it to clear my doubt?

  • $\begingroup$ Are you asking for the value of $3^{0.7}$ or are you asking how to interpret it? $\endgroup$ – user307169 Jul 4 '16 at 19:01
  • $\begingroup$ $0.7=\frac 7{10}$, so $3^{0.7}=3^{\frac 7{10}} = (3^7)^{\frac 1{10}}$. You can use this approach for any $a^b$ where $a>0$ and $b$ a rational number. Then if $b$ is irrational, you can uniquely determine $a^b$ due to the fact that $\Bbb Q$ is dense in $\Bbb R$. $\endgroup$ – user137731 Jul 4 '16 at 19:05
  • $\begingroup$ $$x^{\frac p q} = (x^p)^{\frac 1 q} = \sqrt[q]{x^p}$$ $\endgroup$ – Rodrigo de Azevedo Jul 4 '16 at 19:06
  • $\begingroup$ The tenth root of $3^7=2187$. $\endgroup$ – Bernard Jul 4 '16 at 19:09
  • $\begingroup$ But it is just conversion of equation which does not solve my confusion. Can you please say something like: 3^2 means multiple value of 2 times 3 $\endgroup$ – Asif Iqbal Jul 4 '16 at 19:12

Exponents less than 1 are better understood in terms of fractions rather than decimals:

What value, when multiplied by itself, is equal to 3? The only possibility is ±$\sqrt{3}$. This is why we understand an exponent of $\frac{1}{2}$ to be equivalent to the square root. Similarly, an exponent of $\frac{1}{3}$ is the cube root. Any value in the numerator is the equivalent of an integer exponent.

To determine the value of $3^{0.7}$, convert the exponent to a fraction: $0.7=\frac{7}{10}$. So $3^{0.7}=3^\frac{7}{10}=(3^7)^{1/10}=\sqrt[10]{3^7}≈2.16$.

  • $\begingroup$ Nah, $-\sqrt{3}$ is also 3 when squared, but not if you consider negative numbers to be unreal. $\endgroup$ – mathreadler Jul 4 '16 at 19:20
  • $\begingroup$ @mathreader Thanks for the suggested edit. :) $\endgroup$ – scott Jul 4 '16 at 21:48

Intuitively, the way to think about exponentiation is by considering exponential growth/ decay.

Let's say we have a bacterial sample growing on an agar medium. And we happen to know that this bacteria will triple in population size every minute.

enter image description here

Then, if $P_0$ is the initial population, the population at any later time will be $$P(t) = P_0\cdot 3^t$$ Let's make sure that's correct. What do we get at $t=0$? We get $P(0)=P_0\cdot 3^0 = P_0\cdot 1 = P_0$. OK, that's good. What about $t=1$? Then we get $P(1) = P_0\cdot 3^1 = 3P_0$: i.e. triple the population. What about $t=2$? Do we get triple triple (i.e. $9$ times) the population? Yes we do.

enter image description here

So then what does $3^{0.7}$ mean? It's the ratio of the amount of the bacterial cells after $0.7$ minutes to the initial amount -- $\dfrac{P(0.7)}{P_0}$ -- given the population triples every minute.


Note that $0.7 = \dfrac{7}{10}$, so your specific example can be thought of as:

$$ 3^{0.7} = 3^{7/10} = \left(3^7\right)^{1/10} $$

Here we can think of $3^7$ the same way you thought of $3^4$, etc. And the $1/10$ can be thought of as "the tenth root."

This reasoning can be applied for any rational exponent. Irrational exponents are more complicated. Typically, irrational exponents are rigorously justified in an undergraduate real analysis course.


It's easy to understand when the exponent is a natural number; it's also easy to understand when the reciprocal of the exponent is a natural number. For example, $3^{1/10}=\sqrt[10]{3}$ is the positive number such that, when you multiply it by itself ten times, you get $3$.

Further, $(a^{b})^c = a^{bc}$. Putting these together, $3^{0.7}=(3^{7})^{1/10} = (\sqrt[10]{3^7})$.This process works for any rational number $\frac{p}{q}$, as $x^{p/q} = \sqrt[q]{x^p}$. For irrational exponents, one can approximate the value by using successively better rational approximations for the exponent.

This is a bit more complicated, but alternatively, one can define $a^b =e^{b\ln(a)}$, where $$\ln(x) = \int_1^x\frac{1}{t}dt$$ and $e^x$ is the inverse of $\ln(x)$.


Take the definition of exponents in $ \mathbb N $, now observe how,

$$ (a^x)^y = a^{xy} $$

To preserve this identity, logically,

$$ a^{\frac{x}{y}} = a^x \sqrt[y]{a} $$

To extend this to $ x \in \mathbb R $, use the Cauchy sequence, $ C_n(x) $, defining $ x $,

$$ a^x = \lim_{n \to \infty} a^{C_n(x)} $$


Not the answer you're looking for? Browse other questions tagged or ask your own question.