What is the answer to 2^{2x}+2^x-42=0 ?

The answer to 2^{2x}+2^x-42=0 is x=(ln(6))/(ln(2))

Solve for x: 2^{2x}+2^x-42=0

The solution is x=(ln(6))/(ln(2))

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2^{2 x} + 2^{x} - 42 = 0

x = \frac{ln ( 6 )}{ln ( 2 )}

Decimal Notation

x = 2.58496 \dots

Solution steps

2^{2 x} + 2^{x} - 42 = 0

Apply exponent rules

(2^{x})^{2} + 2^{x} - 42 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} + u - 42 = 0

Solve

u^{2} + u - 42 = 0 : u = 6, u = - 7

u = 6, u = - 7

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 6 : x = \frac{ln ( 6 )}{ln ( 2 )}

Solve

2^{x} = - 7 :

No Solution for

x \in R

x = \frac{ln ( 6 )}{ln ( 2 )}

6^{x + 1} = 216

e^{\frac{5 y - 3}{4}} = e^{\frac{4 + y}{3}}

0. 5^{x + 7} \times 0. 5^{1 - 2 x} = 2

x^{6} + 5 x^{3} + 6

y, x = y^{3}

What is the answer to 2^{2x}+2^x-42=0 ?
The answer to 2^{2x}+2^x-42=0 is x=(ln(6))/(ln(2))
Solve for x: 2^{2x}+2^x-42=0
The solution is x=(ln(6))/(ln(2))