${ {e}^{x} }^{2} = \frac{1}{ {e}^{2} } \times {e}^{3x} \\ \Leftrightarrow { {e}^{x} }^{2} = {e}^{ 3x} \times {e}^{ - 2} \\ \Leftrightarrow { {e}^{x} }^{2} = {e}^{3x - 2} \\ \Leftrightarrow {x}^{2} = 3x - 2 \\ \Leftrightarrow {x}^{2} - 3x + 2 = 0 \\ \Leftrightarrow (x - 1)(x - 2) = 0 \\ x = 1 \: \vee \: x = 2 \\ \\ {32}^{ {x}^{2} - 2x } = \frac{1}{ {4}^{x} } \\ \Leftrightarrow {( {2}^{5} )}^{ {x}^{2} - 2x } = {2}^{ - 2x} \\ \Leftrightarrow {2}^{5 {x}^{2} - 10x } = {2}^{ - 2x} \\ \Leftrightarrow 5 {x}^{2} - 10x = - 2x \\ \Leftrightarrow 5 {x}^{2} - 8x = 0 \\ \Leftrightarrow x(5x - 8) = 0 \\ \Leftrightarrow x = 0 \: \vee \: x = \frac{8}{5} \\ \\ { {e}^{x} }^{3} = {2}^{2x} \\ \Leftrightarrow {x}^{3} = ln( {2}^{2x} ) \\ \Leftrightarrow {x}^{3} - 2x ln(2) = 0 \\ \Leftrightarrow x( {x}^{2} - 2 ln(2) ) = 0 \\ \Leftrightarrow x = 0 \:\vee \: {x}^{2} = 2 ln(2) \\ \Leftrightarrow x = 0 \:\vee \: x = \sqrt{2 ln(2) } \:\vee \: x = - \sqrt{2 ln(2) }$

8 0

4 years ago

Please help!!!!!! It's pythagoras!!!!!!

mote1985 [20]

The answer to your problem A=60

3 0

4 years ago

____ < √97 < ____ integers to complete the following inequality.

< √97 < integers to complete the following inequality.