What is the answer to 4^{2x}-2^{2x+1}=35 ?

The answer to 4^{2x}-2^{2x+1}=35 is x=(ln(7))/(2ln(2))

Solve for x: 4^{2x}-2^{2x+1}=35

The solution is x=(ln(7))/(2ln(2))

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4^{2 x} - 2^{2 x + 1} = 35

x = \frac{ln ( 7 )}{2 ln ( 2 )}

Solution steps

4^{2 x} - 2^{2 x + 1} = 35

Apply exponent rules

(2^{x})^{4} - (2^{x})^{2} \cdot 2^{1} = 35

Rewrite the equation with

2^{x} = u

(u)^{4} - (u)^{2} \cdot 2^{1} = 35

Solve

u^{4} - u^{2} \cdot 2^{1} = 35 : u = 7, u = - 7

u = 7, u = - 7

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 7 : x = \frac{ln ( 7 )}{2 ln ( 2 )}

Solve

2^{x} = - 7 :

No Solution for

x \in R

x = \frac{ln ( 7 )}{2 ln ( 2 )}

\frac{2 5 ^{x} + 125}{6} = 5^{x + 1}

1 0^{x} = 200

8 (2^{2 x}) - 65 (2^{x}) + 8 = 0

1000000 = 2^{n} \cdot 5^{n}

6 4^{x} - 8 = 7 (4^{x} - 2)

What is the answer to 4^{2x}-2^{2x+1}=35 ?
The answer to 4^{2x}-2^{2x+1}=35 is x=(ln(7))/(2ln(2))
Solve for x: 4^{2x}-2^{2x+1}=35
The solution is x=(ln(7))/(2ln(2))