Question

find the value of 7 raise to power x ​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$= \sqrt{7 \sqrt{7 \sqrt{7 \sqrt{7 \sqrt{7} } } } } = {7}^{x} \\ = \sqrt{7 \sqrt{7 \sqrt{7 \sqrt{7 \times {7}^{ \frac{1}{2} } } } } } = {7}^{x} \\ = > \sqrt{7 \sqrt{7 \sqrt{7 \times {7}^{ \frac{1}{2} \times {7}^{ \frac{1}{4} } } } } } = {7}^{x} \\ = \sqrt{7 \sqrt{7 \times {7}^{ \frac{1}{2} \times {7}^{ \frac{1}{4} \times {7}^{ \frac{1}{8} } } } } } = {7}^{x} \\ = \sqrt{7 \times {7}^{ \frac{1}{2} \times {7}^{ \frac{1}{4} \times {7}^{ \frac{1}{8} \times {7}^{ \frac{1}{16} } } } } } = {7}^{x} \\ = {7}^{ {1}^{2} } \times {7}^{ \frac{1}{4} } \times {7}^{ \frac{1}{8} } \times {7}^{ \frac{1}{16} } \times {7}^{ \frac{1}{32} } = {7}^{x} \\ = > {7}^{ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} } = {7}^{x} \\ = > \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \: are \: in \: gp \\ = > a = \frac{1}{2} \: r = \frac{ \frac{1}{4} }{ \frac{1}{2} } = \frac{1}{2} \\ = > sn = \frac{a \times (1 - {r}^{n} }{1 - r} \\ = > s5 = \frac{ \frac{1}{2} \times (1 - { \frac{1}{2} }^{5} }{1 - \frac{1}{2} } \\ = > s5 = \frac{31}{16} \\ = > {7}^{ \frac{31}{16} } = {7}^{x} \\ = > x = \frac{31}{16}$