Question

ALGUIEN QUE ME AYUDE CON ESO :( X FA

Answer 1

Hola!

Explicación paso a paso:

■ Solución:

a)

${2}^{3x - 5} = 1024 \\ \\ {2}^{3x - 5} = {2}^{10} \\ \\ 3x - 5 = 10 \\ \\ 3x = 15 \\ \\ x = 5$

b)

$\sqrt{ {4}^{ - 2x + 6} } = \frac{1}{8} \\ \\ {4}^{ - 2x + 6} = {( \frac{1}{8} )}^{2} \\ \\ {4}^{( - 1)(2x - 6)} = \frac{1}{64} \\ \\ {( {4}^{2x - 6} )}^{ - 1} = {(64)}^{ - 1} \\ \\ {4}^{2x - 6} = 64 \\ \\ {4}^{2x - 6} = {4}^{3} \\ \\ 2x - 6 = 3 \\ \\ 2x = 9 \\ \\ x = \frac{9}{2}$

c)

${32}^{ {x}^{2} - 5x - 3 } = 1 \\ \\ {32}^{ {x}^{2} - 5x - 3 } = {32}^{0} \\ \\ {x}^{2} - 5x - 3 = 0 \\ \\ \frac{ - ( - 5) \frac{ + }{ - } \sqrt{ {( - 5)}^{2} - 4(1)( - 3)} }{2(1)} \\ \\ \frac{5 \frac{ + }{ - } \sqrt{25 + 12} }{2} \\ \\ \frac{5 \frac{ + }{ - } \sqrt{37} }{2} \\ \\ \: \: \: \: \: \: \: \frac{5 + \sqrt{37} }{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{5 - \sqrt{37} }{2}$

d)

${25}^{x - 4} = {( \frac{1}{5} )}^{2x} \\ \\ {25}^{( - 1)(4 - x)} = {( {5}^{ - 1} )}^{2x} \\ \\ {( {25}^{4 - x}) }^{ - 1} = {( {5}^{2x} )}^{ - 1} \\ \\ {25}^{4 - x} = {5}^{2x} \\ \\ {( {5}^{2}) }^{4 - x} = {5}^{2x} \\ \\ {5}^{2(4 - x)} = {5}^{2x} \\ \\ {5}^{8 - 2x} = {5}^{2x} \\ \\ 8 - 2x = 2x \\ \\ 8 = 4x \\ \\ 2 = x \\ \\ x = 2$

e)

${2}^{x + 1} + {2}^{x} + {2}^{x - 1} = 112 \\ \\ {2}^{x} . {2}^{1} + {2}^{x} + {2}^{x} . {2}^{ - 1} = 112 \\ \\ {2}^{x} ( {2}^{1} + 1 + {2}^{ - 1} ) = 112 \\ \\ {2}^{x} (2 + 1 + \frac{1}{2} ) = 112 \\ \\ {2}^{x} (3 + \frac{1}{2} ) = 122 \\ \\ {2}^{x} ( \frac{7}{2} ) = 122 \\ \\ {2}^{x} = 112. \frac{2}{7} \\ \\ {2 }^{x} = \frac{224}{7} \\ \\ {2}^{x} = 32 \\ \\ {2}^{x} = {2}^{5} \\ \\ x = 5$

f)

${9}^{x - 1} - 2 . {3}^{x} - 27 = 0 \\ \\ {( {3}^{2} )}^{x - 1} - 2. {3}^{x} = 27\\ \\ {3}^{2x - 2} - 2. {3}^{x} = 27\\ \\ {3}^{2x} . {3}^{ - 2} - 2. {3}^{x} = {3}^{3} \\ \\ {3}^{x} ( {3}^{x} . {3}^{ - 2} - 2) = {3}^{3} \\ \\ x = 3$