Question

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Answer 1

Verified answer

Respuesta:

[tex] {4}^{x + 1} = \frac{ {( {( {4}^{2} )}^{3} )}^{0} \times {4}^{2} }{5 - {(4}^{9} )^{0} } \\ {4}^{x + 1} = \frac{ {4}^{2 \times 3 \times 0} \times {4}^{2} }{5 - {4}^{9 \times 0} } \\ {4}^{x + 1} = \frac{1 \times {4}^{2} }{5 - 1} \\ {4}^{x + 1} = \frac{ {4}^{2} }{4} \\ {4}^{x + 1} = {4}^{2 - 1} \\ x + 1 = 2 - 1 \\ x = 0[/tex]

[tex] \frac{ ({(8)}^{2})^{3} \times {(((8)}^{8})^{3})^{0} \times \sqrt{64} }{ \sqrt[3]{512} } = {8}^{x} \\ \frac{{8}^{2 \times 3} \times {8}^{8 \times 3 \times 0} \times 8}{8} = {8}^{x} \\ \frac{ {8}^{6} \times 1 \times 8 }{8} = {8}^{x} \\ {8}^{7 - 1} = {8}^{x} \\ 6 = x[/tex]