Solve the equation $2^{2x} - 9(2^x) + 20 = 0$.
Understand the Problem
The question asks us to solve an exponential equation. To solve it, we can simplify the equation using laws of exponents, substitute a variable, solve for the variable, and then use logarithms to finally find the value of x.
Answer
$x = 2$ or $x = \frac{\ln{5}}{\ln{2}}$
Answer for screen readers
$x = 2$ or $x = \frac{\ln{5}}{\ln{2}}$
Steps to Solve
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Rewrite the equation
We can rewrite $2^{2x}$ as $(2^x)^2$. This gives us:
$(2^x)^2 - 9(2^x) + 20 = 0$
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Substitute a variable
Let $y = 2^x$. Substituting $y$ into the equation, we get a quadratic equation:
$y^2 - 9y + 20 = 0$
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Solve the quadratic equation
We can factor the quadratic equation:
$(y - 4)(y - 5) = 0$
So the solutions for $y$ are:
$y = 4$ or $y = 5$
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Substitute back and solve for x
Since $y = 2^x$, we have two equations to solve for $x$:
$2^x = 4$ or $2^x = 5$
For $2^x = 4$, we can write $4$ as $2^2$, so $2^x = 2^2$. Therefore:
$x = 2$
For $2^x = 5$, we can use logarithms to solve for $x$:
$x = \log_2{5}$
We can also write this using the change of base formula as:
$x = \frac{\ln{5}}{\ln{2}}$
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Final Answer
The solutions for $x$ are $x=2$ or $x = \frac{\ln{5}}{\ln{2}}$.
$x = 2$ or $x = \frac{\ln{5}}{\ln{2}}$
More Information
The solutions to the equation $2^{2x} - 9(2^x) + 20 = 0$ are $x=2$ and $x = \log_2{5} \approx 2.3219$. These values of $x$ make the original equation equal to zero.
Tips
A common mistake is to forget to substitute back to find $x$ after solving for $y$. Another mistake is incorrectly applying logarithm rules when solving for $x$ in the equation $2^x = 5$.
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