Question

solve this sum. Fast​

Answer 1

4 ^ x - 5 × 2^x + 4 =0

Let 2^x = a

a^2 -5×a +4 =0

a^2 -4a - a +4 =0

a(a- 4) -1( a-4) =0

(a-1)(a-4) = 0

a = 1 or 4

2^x = 1 or 2^x = 4

××××××× here equating like terms

x =2

Answer 2

Step-by-step explanation:

${4}^{x} - 5. {2}^{x} + 4 = 0 \\ = > {( {(2)}^{2} )}^{x} - 5. {2}^{x} + 4 = 0 \\ = > {2}^{2x} - 5 .{2}^{x} + 4 = 0 \\ = > {2}^{x}. {2}^{x} - 5. {2}^{x} + 4 = 0 \\ let \: {2}^{x } \: be \: y \\ = > y.y - 5y + 4 = 0 \\ = > {y}^{2} - 5y + 4 = 0 \\ = > {y}^{2} - 4y - y + 4 = 0 \\ = > y(y - 4) - 1(y - 4) = 0 \\ = > (y - 4)(y - 1) = 0$

Therefore,

either. or

y - 4 = 0. y - 1 = 0

=> y = 4. => y = 1

Therefore y = 4 or 1

Case I:

$y=4\\=>{2}^{x}=4\\=>{2}^{x}={2}^{2}$

if bases are equal then power are also equal

$=>x=2$

_________________________

Case II:

$y=1\\=>{2}^{x}=1\\=>{2}^{x}={2}^{0}$

if bases are equal then power are also equal

$=>x=0$

_____________________________

So, required value of x is 2 or 0

__________________________________________

//Hope this will Helped You //

//Please mark it as Brainliest//