What is the answer to 4^{(x)}-12*2^{(x)}+35=0 ?

The answer to 4^{(x)}-12*2^{(x)}+35=0 is x=(ln(7))/(ln(2)),x=(ln(5))/(ln(2))

Solve for x: 4^{(x)}-12*2^{(x)}+35=0

The solution is x=(ln(7))/(ln(2)),x=(ln(5))/(ln(2))

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4^{(x)} - 12 \cdot 2^{(x)} + 35 = 0

x = \frac{ln ( 7 )}{ln ( 2 )}, x = \frac{ln ( 5 )}{ln ( 2 )}

Decimal Notation

x = 2.80735 \dots, x = 2.32192 \dots

Solution steps

4^{(x)} - 12 \cdot 2^{(x)} + 35 = 0

Apply exponent rules

(2^{x})^{2} - 12 \cdot 2^{x} + 35 = 0

Rewrite the equation with

2^{x} = u

(u)^{2} - 12 u + 35 = 0

Solve

u^{2} - 12 u + 35 = 0 : u = 7, u = 5

u = 7, u = 5

Substitute back

u = 2^{x},

solve for

x

Solve

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

Solve

2^{x} = 5 : x = \frac{ln ( 5 )}{ln ( 2 )}

x = \frac{ln ( 7 )}{ln ( 2 )}, x = \frac{ln ( 5 )}{ln ( 2 )}

x 143 = 12 e^{1.4 x + 1.3}

a^{a} = 27

63 = 1 3^{x}

2^{x} + 2^{x - 1} = 12

3^{2 x - 2} = 3

What is the answer to 4^{(x)}-12*2^{(x)}+35=0 ?
The answer to 4^{(x)}-12*2^{(x)}+35=0 is x=(ln(7))/(ln(2)),x=(ln(5))/(ln(2))
Solve for x: 4^{(x)}-12*2^{(x)}+35=0
The solution is x=(ln(7))/(ln(2)),x=(ln(5))/(ln(2))