Question

Solve the equation for x​

Answer 1

Given

$\sf \to \: {4}^{x - 1} \times (0.5) {}^{3 - 2x} = \bigg(\dfrac{1}{8} \bigg) ^{x}$

To find The value of x

$\sf \to \: {4}^{x - 1} \times \bigg( \dfrac{1}{2} \bigg) {}^{3 - 2x} = \bigg(\dfrac{1}{8} \bigg) ^{x}$

$\sf \to \: {2}^{2(x - 1)} \times \bigg( \dfrac{1}{2} \bigg) {}^{3 - 2x} = \bigg(\dfrac{1}{8} \bigg) ^{x}$

$\sf \to \: {2}^{2x - 2} \times {2}^{ - 1(3 - 2x)} = {8}^{ - 1(x)}$

$\sf \to \: {2}^{2x - 2} \times {2}^{2x - 3} = {2}^{ - 3x}$

$\sf \to \: {2}^{2x - 2 + 2x - 3} = {2}^{ - 3x}$

$\sf \to \: {2}^{4x - 5} = {2}^{ - 3x}$

$\sf \to \: 4x - 5 = - 3x$

$\sf \to \: 4x + 3x = 5$

$\sf \to \: 7x = 5$

$\sf \to \: x = \dfrac{5}{7}$

More Information

Some properties of exponential

$\sf \to \: \dfrac{1}{a} = {a}^{ - 1}$

$\sf \to \: {a}^{0} = 1$

$\sf \to \: ( {a}^{m} )^{n} = {a}^{mn}$

$\sf \to \bigg( \dfrac{a^{m} }{ {a}^{n} } \bigg) = {a}^{m - n}$

$\sf \to \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

Answer 2

$\huge{\underbrace{\text{Question}}}$

$\sf{\large{4 {}^{x - 1} \times (0.5) {}^{3 - 2x} = { (\frac{1}{8} }^{x}) }}$

Solve the equation for x

$\huge{\underbrace{\text{\gray{Answer}}}}$

$\rightarrow{\sf{4 {}^{x - 1} \times (0.5) {}^{3 - 2x} = ({ \frac{1}{8}) }^{x} }}$

$\sf{\rightarrow{ { {2}^{2} }^{(x - 3)} \times { \frac{1}{2} }^{3 - 2x} = (\frac{1}{ {2}^{3} } }) {}^{x} }$

$\rightarrow{\sf{ {2}^{2x - 2} \times 2 {}^{ - 1} {}^{(3 - 2x)} = 2 {}^{ - 3} {}^{(x)} }}$

$\sf{\rightarrow{2 {}^{2x - 2} \times {2}^{ - 3 + 2x} = { 2 }^{ - 3x} }}$

$\sf{\rightarrow{2 {}^{2x - 2 + ( - 3 + 2x)} = {2}^{ - 3x} }}$

$\sf{\rightarrow{2 {}^{4x - 5} = {2}^{ - 3x} }}$

$\sf{\therefore{4x - 5 = - 3x}} \: \: \: \: \boxed{\sf{\blue{\because{ {a}^{x} = {a}^{y} = > x = y}}}}$

$\sf{\rightarrow{4x + 3x = 5}}$

$\sf{\rightarrow{7x = 5}}$

${\boxed{\sf{\therefore{\orange{x = \frac{5}{7} }}}}}$

$\underline{\boxed{\rm{Hence, \: the\:answer\:is\:\bold{x = \frac{5}{7} }}}}$

$\large{\underline{\underline{\sf{\purple{More\:information}}}}}$

$\sf{\rightarrow{\bold{ {a}^{x} + {a}^{y}}}} = {a}^{xy} \\ \\ \sf{\rightarrow{\bold{ {a}^{x} \times {a}^{y}}}} = {a}^{x + y} \\ \\ \sf{\rightarrow{\bold{ {a}^{x} - {a}^{y}}}} = {a}^{x \div y} \\ \\ \sf{\rightarrow{\bold{ {a}^{x} \div {a}^{y}}}} = {a}^{x - y} \\ \\ \sf{\rightarrow{\bold({ {a}^{x} ) {}^{y} }}} = {a}^{x y}$