Question

Розв^яжіть рівняння, вправа 2.1.

Answer 1

Ответ:

${4}^{x} = 64 \\ {4}^{x} = {4}^{3} \\ x = 3$

${3}^{x} = \frac{1}{81} \\ {3}^{x} = {3}^{ - 4} \\ x = - 4$

${0.6}^{2x - 3} = 1 \\ {0.6}^{2x - 3} = {0.6}^{0} \\ 2x - 3 = 0 \\ 2x = 3 \\ x = \frac{3}{2}$

${10}^{ - x} = 0.001 \\ {10}^{ - x} = {10}^{ - 3} \\ - x = - 3 \\ x = 3$

${2}^{5 - x} = {2}^{3x - 7} \\ 5 - x = 3x - 7 \\ - x - 3x = - 7 - 5 \\ - 4x = - 12 \\ x = 12 \div 4 \\ x = 3$

${8}^{x} = 16 \\ {2}^{3x} = {2}^{4} \\ 3x = 4 \\ x = \frac{4}{3}$

$\sqrt{ {5}^{x} } = 25 \\ {5}^{ \frac{1}{2}x } = {5}^{2} \\ \frac{1}{2} x = 2 \\ x = 2 \div \frac{1}{2} \\ x = 4$

${0.25}^{ {x}^{2} - 4 } = {2}^{ {x}^{2} + 1 } \\ {2}^{ - 2( {x}^{2} - 4) } = {2}^{ {x}^{2} + 1} \\ - 2( {x}^{2} - 4 )= {x}^{2} + 1 \\ - 2 {x}^{2} + 8 = {x}^{2} + 1 \\ - 2 {x}^{2} - {x}^{2} = 1 - 8 \\ - 3 {x}^{2} = - 7 \\ {x}^{2} = \frac{3}{7} \\ x.1.2 .= - + \sqrt{ \frac{3}{7} }$