Write the fraction 81/16 as a power. Which base and exponent work, and is 81/16 a power of any whole number?

Step-by-step solution

1. 1

   Factor the numerator and denominator into prime powers. Repeated division gives

   $81=3\cdot 27=3\cdot 3\cdot 9=3^{4},\qquad 16=2\cdot 8=2\cdot 2\cdot 4=2^{4}.$

   The point of factoring first is to see whether the two exponents match — that is exactly the condition for the fraction to be a single power.

2. 2

   Use the quotient-of-powers identity in reverse. Because both exponents are $4$, the law $\dfrac{a^{n}}{b^{n}}=\left(\dfrac{a}{b}\right)^{n}$ applies directly:

   $\frac{81}{16}=\frac{3^{4}}{2^{4}}=\left(\frac{3}{2}\right)^{4}.$

   Most students know this identity left-to-right; reading it right-to-left is what turns a fraction into a power.

3. 3

   Check the exponent cannot be reduced. Could $\tfrac{81}{16}$ also be written with a smaller exponent, say as a square? That would need $\left(\tfrac{3}{2}\right)^{4}=t^{2}$ with $t=\left(\tfrac{3}{2}\right)^{2}=\tfrac94$, so indeed

   $\frac{81}{16}=\left(\frac{9}{4}\right)^{2}=\left(\frac{3}{2}\right)^{4}.$

   Both forms are correct; $\left(\tfrac32\right)^{4}$ is the one with the smallest base, and the exponent $4$ is the largest possible because $3$ and $2$ are prime.

4. 4

   Answer the "power of a whole number" part. As a decimal $\tfrac{81}{16}=5.0625$, which is not an integer, so it cannot equal $m^{k}$ for whole numbers $m$ and $k\ge 1$. The base is necessarily fractional — a fraction in lowest terms is an integer power only if the denominator is $1$.

5. 5

   Verify exactly, not by decimals. Computing $\left(\tfrac32\right)^{4}$ with fractions: $\tfrac32\cdot\tfrac32=\tfrac94$, and $\tfrac94\cdot\tfrac94=\tfrac{81}{16}$ ✓. Doing the check in exact fractions rather than with $1.5^{4}=5.0625$ avoids any rounding doubt.