Formulas for Fractional Exponents $\frac pq$ with Logarithms

Derivation of Formulas $\def\sign{\hbox{sign}}$

We will talk about the calculation of $b^\frac{p}{q}$ in the case where p and q are relatively prime integers. We will see that the latter is defined for all real b when q is odd, but it only defined for b non-negative when q is even. By convention, that is, by definition, $x = b^\frac{p}{q}$ when and only when \[x^q= b^p \]

This definition depends on the existence of solutions. In the case where q is odd, the sign of $x^q$ and $x$, and $x^q$ may take any real value. In the case where q is even, the the sign of $x^q$ is positive or 0. For $b^p$ to be in the range of $f(x) = x^q$, p must be even as well, or if its odd, b must be zero or positive. So some restrictions apply. Using the aid of logarithms, exponentials and sign function, we will find logarithmic-exponential formulas for the solution or solutions of \[x^q= b^p \] The latter will provide formulas for $ b^\frac{p}{q}$ and if present, another solution of $x^q= b^p$ Details follow.

Solving $x^q= b^p$ when b is positive

We look for a positive solution x. In that case, both sides of the equation \[x^q= b^p \] are positive and hence in the domain of the natural logarithm. Taking logarithms of both sides gives $ \ln(x^q) =\ln(b^p)$. That implies first \[ q \cdot \ln(x ) = p \cdot \ln(b ) \] then \[ \ln(x ) = \frac pq \cdot\ln(b ) \] Finally $x = \exp(\ln(x))$ implies \begin{eqnarray*} x &=& \exp(\ln(x)) \\ &=& \exp\left(\frac pq \cdot\ln(b )\right) \\ \end{eqnarray*} So the latter is what x must be if x is positive. The properties of the natural logarithms and its inverse, the exponential function, imply the latter gives a positive solution of $x^q= b^p.$ Exercise Check that.

The foregoing considerations give \[b^{\frac pq } =\exp\left(\frac pq \cdot\ln(b )\right) \] whenever b is positive and $\frac pq$ is any rational number. For equivalent fractions, the property \[\frac PQ = \frac pq \] implies \[ b^{\frac pq } =b^{\frac PQ} \] In the case p = 1, the number \[x = b^{\frac 1q } =\exp\left(\frac 1q \cdot\ln(b )\right) \] satisfies $x^q=1$. Hence \[ x = b^{\frac 1q }= \root q \of b \].

Now when both x and y are nonzero, $x^q= b^p$ implies \[|x|^q = |b|^p \]

In consequence \begin{eqnarray*} \ln(|x|^q) &= & \ln(|b|^p) \\ \mbox{or} \\ q \ln(|x| ) &= & p\ln(|b|) \end{eqnarray*} The latter implies \begin{eqnarray*} \ln(|x|) &= & \frac pq \ln(|b| ) \\ \mbox{and hence} \\ |x| &= & \exp \left(\frac pq \ln(|b| ) \right) \\ &=& \exp \left(\frac 1q \ln(|b|^p ) \right) \end{eqnarray*}

It easily seen that |x| &= & \pm \exp \left(\frac pq \ln(|b| ) \right) \\ &=& \exp \left(\frac 1q \ln(|b|^p ) satisfy the equation \[|x|^q = |b|^p \] but with these expressions, there is no guarantee that $x^q= b^p$ regardless of the sign chosen the last formula for x when b is negative, without further sign analysis.

Solving $x^q= b^p$ in the special case q even and nonzero

If x is postive solution of \[x^q= b^p \] then its additive inverse or negative $-x$ is also a solution: Thus for q even and nonzero the equation \[x^q= b^p \] has two roots, a positive root \[x = b^{\frac 1q } =\exp\left(\frac 1q \cdot\ln(b )\right) \] which we call the principal root, and a negative root \[x = -b^{\frac 1q } =-\exp\left(\frac 1q \cdot\ln(b )\right) \] given by the negative of the principal root.

Observe if p is odd and b is negative, there are no solutions x of $x^q= b^p$. For then, equality is not possible as the lefthandside is nonnegative and right hand side is negative.

Solving $x^q= b^p$ when b is zero

The formula for the solution in this case where p and q are non-zero integers with is very simple, \[ 0^{\frac pq } = x = 0 \]

Solving $x^q= b^p$ when b is negative

If p is even the equation $x^q= b^p$ implies $x^q= |b|^p$. From that \[x = \exp\left(\frac pq \cdot\ln(|b|)\right)]

Now assume p is odd to study the remaining cases in which q may be odd or even.

$x^q= b^p$ for p odd and q even, and b negative.

In this situation, b negative implies $b^p$ is negative. However, when q is even, $x^q \ge 0$ for all real numbers x. So the equation $x^q= b^p$ has no solutions. That leaves only the case q odd to study.

$x^q= b^p$ for p odd and q odd, b negative or positive.

Now we consider the case where exponents $p$ and $q$ are both odd, and leave the case where q is even to later. In this both exponents are odd case, $x^q= b^p$ if and only if $(-x)^q= (-b)^p$. From the study of the equation $x^q= b^p$ with b positive, we conclude

\[-x = \exp\left(\frac pq \cdot\ln(-b)\right)\]

and hence

\[ x = \sign(b) \exp\left(\frac pq \cdot\ln(|b|)\right)\]

Here you may take the sign of positive number to be +1 or simply +, the sign of a negative number to -1 or simply -. The foregoing leads to the small both odd theorem.

The foregoing gives \[ b^{\frac pq} = sign(b) \exp\left(\frac pq \cdot\ln(|b|)\right)] when p and q are odd.

Exercises

1. let $f(s)= s^{\frac23}$ otherwise. Graph the function $t = f(s)$ for $|s| \le 8.$ Find its domain and range.

2. Let $f(s)= s^{\frac53}$ otherwise. Graph the function $t = f(s)$ for $|s| \le 32.$ Find its domain and range.

3. let $f(s)= s^{\frac3/2}$ otherwise. Graph the function $t = f(s)$ for $|s| \le 4.$ Find its domain and range.

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
3. Prime Factors - quickly
4. Fractions + Ratios
5. Arith with units - science