Question

2^x+2^x^-1 = 24. find X please do it fast​

Answer 1

Solution:

Given That:

$\rm \longrightarrow {2}^{x} + {2}^{x - 1} = 24$

Can be written as:

$\rm \longrightarrow {2}^{x} + {2}^{x} \cdot {2}^{ - 1} = 24$

Let us assume that:

$\rm \longrightarrow u = {2}^{x}$

So, our equation becomes:

$\rm \longrightarrow u+ u\cdot {2}^{ - 1} = 24$

$\rm \longrightarrow u+ \dfrac{u}{2} = 24$

$\rm \longrightarrow \dfrac{2u + u}{2} = 24$

$\rm \longrightarrow \dfrac{3u}{2} = 24$

$\rm \longrightarrow u= 24 \times \dfrac{2}{3}$

$\rm \longrightarrow u= 16$

Substituting the value of u here, we get:

$\rm \longrightarrow {2}^{x} = 16$

$\rm \longrightarrow {2}^{x} = {2}^{4}$

Comparing base, we get:

$\rm \longrightarrow x = 4$

★ So, the value of x satisfying the given equation is 4.

Learn More:

Laws of exponents.

$\rm 1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\rm 2. \: \: ({a}^{m})^{n} = {a}^{mn}$

$\rm 3. \: \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\rm4. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\rm5. \: \: \bigg(\dfrac{a}{b} \bigg)^{m} = \dfrac{ {a}^{m} }{ {b}^{m} }$

$\rm6. \: \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\rm7. \: \: {a}^{n} = {b}^{n} \rightarrow a = b, n \neq0$

$\rm8. \: \: {a}^{m} = {a}^{n} \rightarrow m = n, a \neq 1$