Solve 3^3*(3^-2+3^-3+3^-4)=

$$3 ^ { 3 } \cdot ( 3 ^ { - 2 } + 3 ^ { - 3 } + 3 ^ { - 4 } ) =$$

Evaluate

$$\frac{13}{3}\approx 4.333333333$$

Solution Steps

Calculate 3 to the power of 3 and get 27.

$$27\left(3^{-2}+3^{-3}+3^{-4}\right)$$

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$27\left(\frac{1}{9}+3^{-3}+3^{-4}\right)$$

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

$$27\left(\frac{1}{9}+\frac{1}{27}+3^{-4}\right)$$

Add \frac{1}{9} and \frac{1}{27} to get \frac{4}{27}.

$$27\left(\frac{4}{27}+3^{-4}\right)$$

Calculate 3 to the power of -4 and get \frac{1}{81}.

$$27\left(\frac{4}{27}+\frac{1}{81}\right)$$

Add \frac{4}{27} and \frac{1}{81} to get \frac{13}{81}.

$$27\times \frac{13}{81}$$

Multiply 27 and \frac{13}{81} to get \frac{13}{3}.

$$\frac{13}{3}$$

Factor

$$\frac{13}{3} = 4\frac{1}{3} = 4.333333333333333$$