Solve {l}{3^{8}*3^{-11}*(-3)^5+(-3)^4}{5^-3*4^-6}

$$\left. \begin{array} { l } { 3 ^ { 8 } \cdot 3 ^ { - 11 } \cdot ( - 3 ) ^ { 5 } + ( - 3 ) ^ { 4 } } \\ { 5 ^ { - 3 } \cdot 4 ^ { - 6 } } \end{array} \right.$$

Sort

$$\frac{1}{512000},72$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 8 and -11 to get -3.

$$sort(3^{-3}\left(-3\right)^{5}+\left(-3\right)^{4},5^{-3}\times 4^{-6})$$

Calculate 3 to the power of -3 and get \frac{1}{27}.

$$sort(\frac{1}{27}\left(-3\right)^{5}+\left(-3\right)^{4},5^{-3}\times 4^{-6})$$

Calculate -3 to the power of 5 and get -243.

$$sort(\frac{1}{27}\left(-243\right)+\left(-3\right)^{4},5^{-3}\times 4^{-6})$$

Multiply \frac{1}{27} and -243 to get -9.

$$sort(-9+\left(-3\right)^{4},5^{-3}\times 4^{-6})$$

Calculate -3 to the power of 4 and get 81.

$$sort(-9+81,5^{-3}\times 4^{-6})$$

Add -9 and 81 to get 72.

$$sort(72,5^{-3}\times 4^{-6})$$

Calculate 5 to the power of -3 and get \frac{1}{125}.

$$sort(72,\frac{1}{125}\times 4^{-6})$$

Calculate 4 to the power of -6 and get \frac{1}{4096}.

$$sort(72,\frac{1}{125}\times \frac{1}{4096})$$

Multiply \frac{1}{125} and \frac{1}{4096} to get \frac{1}{512000}.

$$sort(72,\frac{1}{512000})$$

Convert decimal numbers in the list 72,\frac{1}{512000} to fractions.

$$72,\frac{1}{512000}$$

To sort the list, start from a single element 72.

$$72$$

Insert \frac{1}{512000} to the appropriate location in the new list.

$$\frac{1}{512000},72$$

Evaluate

$$72,\ \frac{1}{512000}$$