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Solve 2^{2x + 1} + 2^{2x} = 3/8.

 Dec 16, 2019
 #1
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Solve for x over the real numbers:
2^(2 x) + 2^(2 x + 1) = 3/8

Simplify and substitute y = 2^(2 x).
2^(2 x) + 2^(2 x + 1) = 3×2^(2 x)
 = 3 y:
3 y = 3/8

Divide both sides by 3:
y = 1/8

Substitute back for y = 2^(2 x):
2^(2 x) = 1/8

Take the logarithm base 2 of both sides:
2 x = -3

Divide both sides by 2:


x = -3/2

 Dec 16, 2019
 #2
avatar+19806 
0

Another method:

 

2(2x+1) =22x *21

Sub in to the original equation

22x * 2   +  22x  = 3/8

22x * 3 = 3/8

22x = 1/8       take log of both sides

2x log 2 = log 1/8

x =  1/2 *  log (1/8) / log2   = -1.5

 Dec 16, 2019
 #3
avatar+4 
+1

It is not so difficult to solve this kind of questions. You just have to follow basic steps

1) Solve for x over the real numbers:
2^(2 x) + 2^(2 x + 1) = 3/8

2) Simplify and substitute y = 2^(2 x).
2^(2 x) + 2^(2 x + 1) = 3×2^(2 x) = 3 y:
3 y = 3/8

3) Divide both sides by 3:
y = 1/8

4) Substitute back for y = 2^(2 x):
2^(2 x) = 1/8

5) Take the logarithm base 2 of both sides:
2 x = -3

6) Divide both sides by 2:
x = -3/2

Now this is the manual calculation which requires some time to solve. If you don't have the concepts and yet want to solve such equations quickly then use online Quadratic Calculator.

 Dec 16, 2019

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