Solve (4/9)^{-2}:(27/8)^{-3}=(frac{2^{2}}{3^{2}})^{-2}:(frac{3^{3}}{2^{3}})^{-3}=[(2/3)^2]^-2:[(3/2)^3]^-3=

$$( \frac { 4 } { 9 } ) ^ { - 2 } : ( \frac { 27 } { 8 } ) ^ { - 3 } = ( \frac { 2 ^ { 2 } } { 3 ^ { 2 } } ) ^ { - 2 } : ( \frac { 3 ^ { 3 } } { 2 ^ { 3 } } ) ^ { - 3 } = [ ( \frac { 2 } { 3 } ) ^ { 2 } ] ^ { - 2 } : [ ( \frac { 3 } { 2 } ) ^ { 3 } ] ^ { - 3 } =$$

Verify

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and -2 to get -4.

$$\frac{\left(\frac{4}{9}\right)^{-2}}{\left(\frac{27}{8}\right)^{-3}}=\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\left(\frac{3}{2}\right)^{3}\right)^{-3}}$$

To raise a power to another power, multiply the exponents. Multiply 3 and -3 to get -9.

$$\frac{\left(\frac{4}{9}\right)^{-2}}{\left(\frac{27}{8}\right)^{-3}}=\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{4}{9} to the power of -2 and get \frac{81}{16}.

$$\frac{\frac{81}{16}}{\left(\frac{27}{8}\right)^{-3}}=\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{27}{8} to the power of -3 and get \frac{512}{19683}.

$$\frac{\frac{81}{16}}{\frac{512}{19683}}=\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Divide \frac{81}{16} by \frac{512}{19683} by multiplying \frac{81}{16} by the reciprocal of \frac{512}{19683}.

$$\frac{81}{16}\times \frac{19683}{512}=\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Multiply \frac{81}{16} and \frac{19683}{512} to get \frac{1594323}{8192}.

$$\frac{1594323}{8192}=\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1594323}{8192}=\frac{\left(\frac{4}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{1594323}{8192}=\frac{\left(\frac{4}{9}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{4}{9} to the power of -2 and get \frac{81}{16}.

$$\frac{1594323}{8192}=\frac{\frac{81}{16}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 3 to the power of 3 and get 27.

$$\frac{1594323}{8192}=\frac{\frac{81}{16}}{\left(\frac{27}{2^{3}}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{1594323}{8192}=\frac{\frac{81}{16}}{\left(\frac{27}{8}\right)^{-3}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{27}{8} to the power of -3 and get \frac{512}{19683}.

$$\frac{1594323}{8192}=\frac{\frac{81}{16}}{\frac{512}{19683}}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Divide \frac{81}{16} by \frac{512}{19683} by multiplying \frac{81}{16} by the reciprocal of \frac{512}{19683}.

$$\frac{1594323}{8192}=\frac{81}{16}\times \frac{19683}{512}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Multiply \frac{81}{16} and \frac{19683}{512} to get \frac{1594323}{8192}.

$$\frac{1594323}{8192}=\frac{1594323}{8192}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Compare \frac{1594323}{8192} and \frac{1594323}{8192}.

$$\text{true}\text{ and }\frac{\left(\frac{2^{2}}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 2 to the power of 2 and get 4.

$$\text{true}\text{ and }\frac{\left(\frac{4}{3^{2}}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 3 to the power of 2 and get 9.

$$\text{true}\text{ and }\frac{\left(\frac{4}{9}\right)^{-2}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{4}{9} to the power of -2 and get \frac{81}{16}.

$$\text{true}\text{ and }\frac{\frac{81}{16}}{\left(\frac{3^{3}}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 3 to the power of 3 and get 27.

$$\text{true}\text{ and }\frac{\frac{81}{16}}{\left(\frac{27}{2^{3}}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate 2 to the power of 3 and get 8.

$$\text{true}\text{ and }\frac{\frac{81}{16}}{\left(\frac{27}{8}\right)^{-3}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{27}{8} to the power of -3 and get \frac{512}{19683}.

$$\text{true}\text{ and }\frac{\frac{81}{16}}{\frac{512}{19683}}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Divide \frac{81}{16} by \frac{512}{19683} by multiplying \frac{81}{16} by the reciprocal of \frac{512}{19683}.

$$\text{true}\text{ and }\frac{81}{16}\times \frac{19683}{512}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Multiply \frac{81}{16} and \frac{19683}{512} to get \frac{1594323}{8192}.

$$\text{true}\text{ and }\frac{1594323}{8192}=\frac{\left(\frac{2}{3}\right)^{-4}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{2}{3} to the power of -4 and get \frac{81}{16}.

$$\text{true}\text{ and }\frac{1594323}{8192}=\frac{\frac{81}{16}}{\left(\frac{3}{2}\right)^{-9}}$$

Calculate \frac{3}{2} to the power of -9 and get \frac{512}{19683}.

$$\text{true}\text{ and }\frac{1594323}{8192}=\frac{\frac{81}{16}}{\frac{512}{19683}}$$

Divide \frac{81}{16} by \frac{512}{19683} by multiplying \frac{81}{16} by the reciprocal of \frac{512}{19683}.

$$\text{true}\text{ and }\frac{1594323}{8192}=\frac{81}{16}\times \frac{19683}{512}$$

Multiply \frac{81}{16} and \frac{19683}{512} to get \frac{1594323}{8192}.

$$\text{true}\text{ and }\frac{1594323}{8192}=\frac{1594323}{8192}$$

Compare \frac{1594323}{8192} and \frac{1594323}{8192}.

$$\text{true}\text{ and }\text{true}$$

The conjunction of \text{true} and \text{true} is \text{true}.

$$\text{true}$$