What is the answer to 2^{x+1}+4^x=80 ?

The answer to 2^{x+1}+4^x=80 is x=3

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2^{x+1}+4^x=80

2^{x + 1} + 4^{x} = 80

x = 3

Solution steps

2^{x + 1} + 4^{x} = 80

Apply exponent rules

2^{x} \cdot 2^{1} + (2^{x})^{2} = 80

Rewrite the equation with

2^{x} = u

u \cdot 2^{1} + (u)^{2} = 80

Solve

u \cdot 2^{1} + u^{2} = 80 : u = 8, u = - 10

u = 8, u = - 10

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 8 : x = 3

Solve

2^{x} = - 10 :

No Solution for

x \in R

x = 3

7^{3 p - 5} = \frac{1}{( 4 9 ^{p} )}

3^{- x - 3} = 4^{- 3 x}

5^{x - 2} \cdot 3^{2 x - 3} = 135

5^{x} = 10

7^{x} + 7^{x - 1} = 8