Question

4*-5.2*+4=0 (* is the power of x) {Ans : 0,2}

Answer 1

$The \: \: answer \: \: is \: \: given \: \: below \\ \\ Given, \: \: {4}^{x} - 5 \times {2}^{x} + 4 = 0 \\ \\ Or, \: \: { {(2}^{2} )}^{x} - 5 \times {2}^{x} + 4 = 0 \\ \\ Or, \: \: {2}^{2x} - 5 \times {2}^{x} + 4 = 0 \\ \\ Or, \: \: {( {2}^{x}) }^{2} - 5 \times {2}^{x} + 4 = 0 \\ \\ Now, \: \: we \: \: consider \: \: {2}^{x} = y \\ \\ Thus ,\: \: we \: \: get \: \: a \: \: new \: \: equation \\ \\ {y}^{2} - 5y + 4 = 0 \\ \\ Or ,\: \: {y}^{2} - (4 + 1)y + 4 = 0 \\ \\ Or, \: \: {y}^{2} - 4y - y + 4 = 0 \\ \\ Or,\: \: y(y - 4) - 1(y - 4) = 0 \\ \\ Or, \: \: (y - 4)(y - 1) = 0 \\ \\ So, \: \: y = 1 \: \: and \: \: y = 4 \\ \\ When \: \: y = 1 ,\\ {2}^{x} = 1 \\ Taking \: \: (log) ,\: \: we \: \: get \\ log ({2}^{x} ) = 1 \\ Or, \: \: x \: log2 = log1 \\ Or ,\: \: x \: log2 = 0 \: \: (Since, \: \: log1 = 0) \\ Or ,\: \: x = 0 \\ \\ Again, \: \: when \: \: y = 4 \\ {2}^{x} = 4\\ Taking \: \:( log ),\: \: we \: \: get \\ log( {2}^{x} ) = log( {2}^{2} ) \\ Or ,\: \: x \: log2 = 2 \: log2 \\ Or, \: \: x = 2 \\ \\ Hence, \: \: the \: \: roots \: \: of \: \: the \: \: given\: \\ equation \: \: are \\ x = 0 \: \: and \: \: x = 2 \\ \\ Thank \: \: you \: \: for \: \: the \: \: question.$

Answer 2

Hey friend, Harish here.

Here is your answer:

Given:

1) An equation 4* - 5.2* + 4 =0

2) * - denotes to the power x.

To find,

The value of x.

Solution:

⇒ $(4)^{x} - 5\times (2^{x}) + 4 = 0$

⇒ $(2^{x})^{2} - 5\times (2^{x}) + 4 = 0$

Let 2ˣ = a.

Then,

⇒  $a^{2} - 5a + 4 = 0$

⇒  $a^{2} - a - 4a +4 =0$

⇒  $a(a-1)-4(a-1) = 0$

⇒  $(a-4)(a-1) =0$

If the equation is 0. Then, a must be equal to 4 (or) 1.

If a =  4.

Then :  2ˣ  = 4 
 
            2ˣ   = 2²  (As the bases are equal we can compare the powers).

Then by comparing we get x = 2.

If a = 1.

Then, 2ˣ = 1 

          2ˣ  = 2⁰ 

Then by comparing we get x = 0.

$\bold{Therefore\ \boxed{x = 0,2}}$
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Hope my answer is helpful to you.