Solve (4/3)^2+7<7*(3/4)^frac{4{2}}

$$( \frac { 4 } { 3 } ) ^ { 2 } + 7 < 7 \cdot ( \frac { 3 } { 4 } ) ^ { \frac { 4 } { 2 } }$$

Verify

Solution Steps

Calculate \frac{4}{3} to the power of 2 and get \frac{16}{9}.

$$\frac{16}{9}+7<7\times \left(\frac{3}{4}\right)^{\frac{4}{2}}$$

Add \frac{16}{9} and 7 to get \frac{79}{9}.

$$\frac{79}{9}<7\times \left(\frac{3}{4}\right)^{\frac{4}{2}}$$

Divide 4 by 2 to get 2.

$$\frac{79}{9}<7\times \left(\frac{3}{4}\right)^{2}$$

Calculate \frac{3}{4} to the power of 2 and get \frac{9}{16}.

$$\frac{79}{9}<7\times \frac{9}{16}$$

Multiply 7 and \frac{9}{16} to get \frac{63}{16}.

$$\frac{79}{9}<\frac{63}{16}$$

Least common multiple of 9 and 16 is 144. Convert \frac{79}{9} and \frac{63}{16} to fractions with denominator 144.

$$\frac{1264}{144}<\frac{567}{144}$$

Compare \frac{1264}{144} and \frac{567}{144}.

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