What are real solutions of exponential equation 4^(3x-2)=2^(x^2+1)?

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4^(3x-2)=2^(x^2+1)

=> 2^(2*(3*x-2)) = 2^(x^2+1)

Taking log of both the sides

=> 6X-4 = X^2 + 1

=>X^2 - 6*X +5 = 0

=>(X-5)(X-1)=0

=> X = 5 or 1.... Answer

We notice that the bases of exponentials are not matching, therefore we'll create matching bases.

`4 = 2^2 =gt 4^(3x-2) = 2^2(3x-2)`

We'll re-write the equation:

`2^2(3x-2) = 2^(x^2+1)`

Since the bases are matching, we'll apply one to one property:

`2(3x-2) = x^2 + 1`

We'll remove the brackets and we'll move all terms to one side:

`6x - 4 - x^2 - 1 = 0`

`-x^2 + 6x - 5 = 0`

`` `x^2 - 6x + 5 = 0`

We'll apply quadratic formula:

x1 = `(6+sqrt(36-20))/2`

x1 = `(6+sqrt16)/2`

x1 = (6+4)/2

x1=5

x2 = (6-4)/2

x2=1

**The real solutions of the given equation are {1 ; 5}.**

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