exponential question

PhantomT828

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Oct 13, 2021
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solving for x in: 9^(2x) - 3^(1 + 2x) + 2 = 0

I tried making the equation 3^(4x) - 3^(1 + 2x) + 2 = 0 , then converting it to a logarithm but I was unable to get an answer.
 
3^(4x) - 3^(1 + 2x) + 2
Hi Phantom. Writing 3^(4x) is good. You could also factor the power 3^(1+2x), to help see the quadratic form that Poliagapitos hinted at.

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

Let us know how it goes.

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solving for x in: 9^(2x) - 3^(1 + 2x) + 2 = 0
I tried making the equation 3^(4x) - 3^(1 + 2x) + 2 = 0 , then converting it to a logarithm but I was unable to get an answer.
Consider [imath] 9^{2x} - 3^{1 + 2x} + 2 = 3x^{4x}-3\cdot 3^{2x}+2=0[/imath]. Now use [imath]y=3^{2x}[/imath].
[imath]y^2-3y+2=(y-2)(y-1)=0[/imath]
Can you finish?
 
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