Easiest way to solve the equation $x^\frac 43 = \frac {16}{81}$

Question

What would be the easiest way to solve this?

$$x^\frac 43 = \frac {16}{81}$$

I saw this in class and have no clue how did they get $$x = \frac 8{27}$$

Answer 1

Remember that

• $(a^x)^y = a^{xy}$

Now raising both sides of your equation to the the power of $\frac{3}{4}$ will preserve the equality and make your life easier,$$x=\big(x^{\frac{4}{3}} \big)^{\frac{3}{4}} = \bigg(\frac{2^4}{3^4}\bigg)^{\frac{3}{4}} = \frac{2^3}{3^3}$$

Answer 2

I'd do it as $\frac {16}{81}=\frac {2^4}{3^4}$, take the fourth root and then cube it. With these power questions, it is often helpful to obtain an explicit factorisation of the numbers involved as the first stage, which will tell you immediately if the exponents are convenient.

Answer 3

Given real number $x$ such that $x^\frac{4}{3} = \frac{16}{81}$:

  $\sqrt[3]{x}^4 = (\frac{2}{3})^4$.

  $\sqrt[3]{x} = \frac{2}{3}$ or $\sqrt[3]{x} = -\frac{2}{3}$.

  $x = (\frac{2}{3})^3$ or $x = (-\frac{2}{3})^3 = -(\frac{2}{3})^3$.

To actively counter the common error made by the other respondents, note all the following for real $x$:

  It is false that $(x^4)^\frac{1}{4} = x$. It is true that $(x^4)^\frac{1}{4} = |x|$.

  It is false that $(x^\frac{4}{3})^\frac{3}{4} = x$. It is true that $(x^\frac{4}{3})^\frac{3}{4} = |x|$.

That is the reason why those respondents do not get the two solutions that I have shown above.

Answer 4

For real exponentiation, see my other answer. For complex exponentiation using the principal branch cut:

Given $x \in \mathbb{C} \smallsetminus \mathbb{R}_{\le 0}$ such that $x^\frac{4}{3} = \frac{16}{81}$:

  $x^\frac{4}{3} = \exp( \frac{4}{3} \ln_π(x) )$ by definition and $\frac{16}{81} = \exp( \ln_π(\frac{16}{81}) ) = \exp( 4 \ln_π(\frac{2}{3}) )$.

  [Note that the last equality is because $\arg_π(\frac{16}{81}) = 0$. Real logarithm rules fail in general!]

  Thus $\frac{4}{3} \ln_π(x) = 4 \ln_π(\frac{2}{3}) + 2πik$ for some integer $k$.

  Thus $\ln_π(x) = 3 \ln_π(\frac{2}{3}) + 2πi\frac{3}{4}k$.

  Thus $\arg_π(x) = 3 \arg_π(\frac{2}{3}) + 2π\frac{3}{4}k = \frac{3}{2}kπ$.

  Thus $k = 0$ because $\arg_π(x) \in (-π,π)$.

  Thus $x = \exp( 3 \ln_π(\frac{2}{3}) ) = (\frac{2}{3})^3$.

As always, one must check that whatever $x$ as found in the last line above is actually a solution to $x^\frac{4}{3} = \frac{16}{81}$. It is, so we have found all solutions where the question is interpreted using the principal branch cut. Using other branch cuts the method is exactly the same but will yield different $k$ and hence different solutions.

Technical note

Some people define the branch cut so that the function is still defined on the negative real line, in which case usually $\arg$ is defined to take values in the half-open interval $(-π,π]$. Under this convention $\ln(-1) = iπ$, whereas under the other convention $\ln(-1)$ is undefined. The main advantage of this convention is that $\ln$ is defined on $\mathbb{C}_{\ne 0}$. The disadvantage is that it is not an open domain, which is useful for other things.

Answer 5

Its same as cubing both sides we get $x^4=\frac{16^3}{81^3}$ thus we have to express it as power of $4$ so we can write it as $\frac{2^{12}}{3^{12}}$ thus taking 4th root $x=8/27$

Answer 6

Method 1: We transform the equation into a difference of fourth powers, then factor to find the roots of the quartic.

\begin{align*} x^{\frac{4}{3}} & = \frac{16}{81}\\ x^4 & = \left(\frac{16}{81}\right)^3\\ x^4 & = \left(\frac{2^4}{3^4}\right)^3\\ x^4 & = \frac{2^{12}}{3^{12}}\\ x^4 & = \left(\frac{2^3}{3^3}\right)^4\\ x^4 & = \left(\frac{8}{27}\right)^4\\ x^4 - \left(\frac{8}{27}\right)^4 & = 0\\ \left[x^2 + \left(\frac{8}{27}\right)^2\right]\left[x^2 - \left(\frac{8}{27}\right)^2\right] & = 0\\ \left[x^2 + \left(\frac{8}{27}\right)^2\right]\left(x + \frac{8}{27}\right)\left(x - \frac{8}{27}\right) & = 0 \tag{1} \end{align*} For any real number $x$, the factor $$x^2 + \left(\frac{8}{27}\right)^2 > 0$$ so the real-valued roots of equation 1 are $$x = \pm \frac{8}{27}$$

Note: If $x$ is a non-zero real number, then $$ (x^n)^{\frac{1}{n}} = \begin{cases} x & \text{if $n$ is odd}\\ |x| & \text{if $n$ is even} \end{cases} $$ Thus, if $x$ is a non-zero real number, we obtain $$\left(x^{\frac{4}{3}}\right)^{\frac{3}{4}} = \left\{\left[\left(x^{\frac{1}{3}}\right)^4\right]^3\right\}^{\frac{1}{4}} = \left\{\left[\left(x^{\frac{1}{3}}\right)^3\right]^4\right\}^{\frac{1}{4}} = (x^4)^{\frac{1}{4}} = |x|$$

Method 2: We raise each side of the equation $x^{\frac{4}{3}} = \frac{16}{81}$ to the $(3/4)$th power, then solve for the real-valued roots. \begin{align*} x^{\frac{4}{3}} & = \frac{16}{81}\\ \left(x^{\frac{4}{3}}\right)^{\frac{3}{4}} & = \left(\frac{16}{81}\right)^{\frac{3}{4}}\\ |x| & = \left[\left(\frac{16}{81}\right)^{\frac{1}{4}}\right]^3\\ |x| & = \left[\left(\frac{2^4}{3^4}\right)^{\frac{1}{4}}\right]^3\\ |x| & = \left(\frac{2}{3}\right)^3\\ |x| & = \frac{8}{27}\\ x & = \pm \frac{8}{27} \end{align*}

Check: If $x = \pm \frac{8}{27}$, then $$\left(\pm \frac{8}{27}\right)^{\frac{4}{3}} = \left[\left(\pm \frac{8}{27}\right)^{\frac{1}{3}}\right]^4 = \left[\left(\pm \frac{2^3}{3^3}\right)^\frac{1}{3}\right]^4 = \left(\pm \frac{2}{3}\right)^4 = \frac{16}{81}$$