What is the answer to 5(2^{2p+1})-17(2^p)+3=0 ?

The answer to 5(2^{2p+1})-17(2^p)+3=0 is p=(ln(3/2))/(ln(2)),p=-(ln(5))/(ln(2))

Solve for p: 5(2^{2p+1})-17(2^p)+3=0

The solution is p=(ln(3/2))/(ln(2)),p=-(ln(5))/(ln(2))

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5 (2^{2 p + 1}) - 17 (2^{p}) + 3 = 0

p = \frac{ln ( \frac{3}{2} )}{ln ( 2 )}, p = - \frac{ln ( 5 )}{ln ( 2 )}

Decimal Notation

p = 0.58496 \dots, p = - 2.32192 \dots

Solution steps

5 (2^{2 p + 1}) - 17 (2^{p}) + 3 = 0

Apply exponent rules

5 (2^{p})^{2} \cdot 2^{1} - 17 \cdot 2^{p} + 3 = 0

Rewrite the equation with

2^{p} = u

5 (u)^{2} \cdot 2^{1} - 17 u + 3 = 0

Solve

5 u^{2} \cdot 2^{1} - 17 u + 3 = 0 : u = \frac{3}{2}, u = \frac{1}{5}

u = \frac{3}{2}, u = \frac{1}{5}

Substitute back

u = 2^{p},

solve for

p

Solve

2^{p} = \frac{3}{2} : p = \frac{ln ( \frac{3}{2} )}{ln ( 2 )}

Solve

2^{p} = \frac{1}{5} : p = - \frac{ln ( 5 )}{ln ( 2 )}

p = \frac{ln ( \frac{3}{2} )}{ln ( 2 )}, p = - \frac{ln ( 5 )}{ln ( 2 )}

e^{e^{x}} = 6

2 7^{\frac{x - 2}{3}} = 81

(\frac{1}{2})^{6 - x} = 2

73 = 1 9^{x}

129 6^{x + 1} = 6^{x + 4}

What is the answer to 5(2^{2p+1})-17(2^p)+3=0 ?
The answer to 5(2^{2p+1})-17(2^p)+3=0 is p=(ln(3/2))/(ln(2)),p=-(ln(5))/(ln(2))
Solve for p: 5(2^{2p+1})-17(2^p)+3=0
The solution is p=(ln(3/2))/(ln(2)),p=-(ln(5))/(ln(2))