Understanding the Exponential Equation 2^(2x+y) = (4^x)(2^y)

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In summary, to solve the equation <i>2^(2x+y) = (4^x)(2^y)</i>, you can use logarithms to isolate the variables. It is also possible to solve the equation without using logarithms by simplifying it to <i>2^(2x+y) = 2^(2x+y)</i>. The possible solutions to this equation depend on the values of <i>x</i> and <i>y</i>, with infinite solutions if both are positive and no real solutions if both are negative. This equation can also be graphed by solving for one variable in terms of the other, and it has applications in areas such as exponential growth and decay and financial calculations.
  • #1
PsychStudent
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Is 22x+y = 4x2y? If I substitute various digits into the x and y variables, it works, but I can't understand why. Can anyone please explain this to me?

For example, if we choose x=3 and y=1,
2(2)(3)+1 = 27 = 128
2(2)(3)+1 = 4321 = 64(2) = 128

Thanks
 
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  • #2
It follows from the basic rules of combining exponents.

[tex] x^a x^b = x^{(a+b)} [/tex]

[tex] x^{ab} = (x^a)^b [/tex]
 
  • #3
I can't believe I didn't figure that out. Thank you!
 

1. How do I solve for the variables in 2^(2x+y) = (4^x)(2^y)?

To solve this equation, you can use logarithms to isolate the variables. Taking the logarithm of both sides, we get: log(2^(2x+y)) = log((4^x)(2^y)). Then using the properties of logarithms, we can simplify the equation to (2x+y)log2 = xlog4 + ylog2. From there, you can solve for either x or y in terms of the other variable.

2. Can this equation be solved without using logarithms?

Yes, it is possible to solve this equation without using logarithms. You can rewrite the equation as 2^(2x+y) = (2^2)^x(2^y) and then simplify to 2^(2x+y) = 2^(2x+y). This shows that both sides of the equation are equal, so any value of x and y that satisfies this equation will be a solution.

3. What are the possible solutions to this equation?

The possible solutions to this equation will depend on the values of x and y. If both x and y are positive, there will be infinite solutions. If x and y are both negative, there will be no real solutions. If one of the variables is positive and the other is negative, there will be no solutions.

4. Can I graph this equation?

Yes, you can graph this equation by first solving for one of the variables in terms of the other. For example, if you solve for y in terms of x, you will get y = (log4 - log2)x. This will give you a linear graph with a slope of (log4 - log2). You can then plot points on the graph to visualize the solutions to the equation.

5. How can this equation be applied in real-life situations?

This equation can be applied in situations where there is exponential growth or decay. For example, it can be used to model the growth of bacteria in a petri dish or the decay of radioactive materials. It can also be used in financial calculations, such as compound interest or population growth. Additionally, this equation is commonly used in mathematics and science to simplify and solve more complex equations.

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