What is the answer to 2^{2x+5}-7(2^x)=0 ?

The answer to 2^{2x+5}-7(2^x)=0 is x=(ln(7)-5ln(2))/(ln(2))

Solve for x: 2^{2x+5}-7(2^x)=0

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

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2^{2 x + 5} - 7 (2^{x}) = 0

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

Decimal Notation

x = - 2.19264 \dots

Solution steps

2^{2 x + 5} - 7 (2^{x}) = 0

Add

7 \cdot 2^{x}

to both sides

2^{2 x + 5} - 7 \cdot 2^{x} + 7 \cdot 2^{x} = 0 + 7 \cdot 2^{x}

Simplify

2^{2 x + 5} = 7 \cdot 2^{x}

Apply exponent rules

(2 x + 5) ln (2) = ln (7) + x ln (2)

Solve

(2 x + 5) ln (2) = ln (7) + x ln (2) : x = \frac{ln ( 7 ) - 5 ln ( 2 )}{ln ( 2 )}

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

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

4 e^{1 + 3 x} - 9 e^{5 - 2 x} = 0

(\frac{1}{3})^{x} = 9

4^{x} + 2^{x + 4} = 3

3^{x + 1} = 27

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