Question

Solve the following quadratic equation
4^x-1-3.2^x-1+2=0​

Answer 1

Answer:

we cab rewrite

4^(x) as 2^(2x):

2^(2x) - 3 * [2^(x) * 2^2] + 32

= 0

2^(2x) - 12 * 2^(x) + 32 = 0

assuming  2^x a value of y:

y^2 - 12y + 32 = 0

Factorising it we gwt

(y - 8)(y - 4) = 0

y = 4, 8

Now substituting 2^x back for y

and solving for each we get

2^x = 4

x = 2

2^x = 8

x = 3

x = 2, 3

Step-by-step explanation:

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Answer 2

Answer:

x = 1 , 2

Solution:

Here,

The given equation is ;

4^(x-1) - 3•2^(x-1) + 2 = 0

The given equation can be rewritten as ;

=> (2²)^(x-1) - 3•2^(x-1) + 2 = 0

=> [2^(x-1)]² - 3•2^(x-1) + 2 = 0

Putting 2^(x-1) = y , the above equation will be reduced to ;

=> y² - 3y + 2 = 0

=> y² - 2y - y + 2 = 0

=> y(y - 2) - (y - 2) = 0

=> (y - 2)(y - 1) = 0

=> y = 1 , 2

Case(1) : If y = 1

=> y = 1

=> 2^(x-1) = 1

=> 2^(x-1) = 2^0

=> x - 1 = 0

=> x = 1

Case(2) : If y = 2

=> y = 2

=> 2^(x-1) = 2^1

=> x - 1 = 1

=> x = 1 + 1

=> x = 2

Hence,

The required answer is ;

x = 1 , 2