Solve for x

Hi Leti,

I am from Scottsdale too.

Let u = 2

^{x}, while u > 02u^2 - u - 3 = (2u-3)(u+1) = 0

So, u = 3/2 = 2

^{x}x = log(base 2) (3/2) <=Answer

Solve for x

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Scottsdale, AZ

Hi Leti,

I am from Scottsdale too.

Let u = 2^{x}, while u > 0

2u^2 - u - 3 = (2u-3)(u+1) = 0

So, u = 3/2 = 2^{x}

x = log(base 2) (3/2) <=Answer

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## Comments

_{Leti, }2

^{(2x+1)}- 2^{x}= 3 ??