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To find the equation of a line that results in three real solutions when intersected with the function [tex]\( k(x) = 2 \log_2 x + 5 \)[/tex], follow these steps:
1. Define the Function:
The function given is [tex]\( k(x) = 2 \log_2 x + 5 \)[/tex].
2. Derivative and Critical Points:
To find points of interest, we calculate the derivative of [tex]\( k(x) \)[/tex]:
[tex]\[ k'(x) = \frac{d}{dx} \left(2 \log_2 x + 5 \right) \][/tex]
Since the derivative of [tex]\( \log_2 x \)[/tex] is [tex]\(\frac{1}{x \ln 2}\)[/tex], we get:
[tex]\[ k'(x) = \frac{2}{x \ln 2} \][/tex]
Setting [tex]\( k'(x) = 0 \)[/tex] to find critical points:
[tex]\[ \frac{2}{x \ln 2} = 0 \][/tex]
This equation has no solutions because the derivative [tex]\(\frac{2}{x \ln 2}\)[/tex] is never zero for [tex]\( x > 0 \)[/tex].
3. Equation of a Line:
Consider a generic line [tex]\( y = mx + b \)[/tex].
4. Form the System of Equations:
Set [tex]\( k(x) \)[/tex] equal to the line:
[tex]\[ 2 \log_2 x + 5 = mx + b \][/tex]
5. Solve for x:
To find solutions for [tex]\( x \)[/tex], rewrite the logarithmic equation:
[tex]\[ 2 \log_2 x + 5 = mx + b \][/tex]
Converting [tex]\(\log_2 x\)[/tex] to natural logarithms ([tex]\(\log_2 x = \frac{\ln x}{\ln 2}\)[/tex]):
[tex]\[ 2 \frac{\ln x}{\ln 2} + 5 = mx + b \][/tex]
6. Analyzing the Solutions:
Rewriting this, we get:
[tex]\[ 2 \ln x + 5 \ln 2 = mx (\ln 2) + b \ln 2 \][/tex]
[tex]\[ 2 \ln x + 5 \ln 2 = mx \ln 2 + b \ln 2 \][/tex]
To find the points of intersection, solve this equation for [tex]\( x \)[/tex]:
[tex]\[ 2 \ln x + 5 \ln 2 = mx \ln 2 + b \ln 2 \][/tex]
This has solutions for [tex]\( x \)[/tex] in terms of the Lambert W function. The solutions are:
[tex]\[ x = -\frac{2 \text{LambertW}(-\sqrt{2} m \sqrt{e^{b \ln 2}} \ln 2 / 16)}{m \ln 2} \text{ and } x = -\frac{2 \text{LambertW}(\sqrt{2} m \sqrt{e^{b \ln 2}} \ln 2 / 16)}{m \ln 2} \][/tex]
These results indicate that under certain conditions for [tex]\( m \)[/tex] and [tex]\( b \)[/tex], the line [tex]\( y = mx + b \)[/tex] can intersect the function [tex]\( k(x) = 2 \log_2 x + 5 \)[/tex] at three points, but it depends heavily on the specific values of [tex]\( m \)[/tex] and [tex]\( b \)[/tex]. Generally, you would select [tex]\( m \)[/tex] and [tex]\( b \)[/tex] to ensure that the logarithmic expression results in real values for the solutions involving the Lambert W function.
1. Define the Function:
The function given is [tex]\( k(x) = 2 \log_2 x + 5 \)[/tex].
2. Derivative and Critical Points:
To find points of interest, we calculate the derivative of [tex]\( k(x) \)[/tex]:
[tex]\[ k'(x) = \frac{d}{dx} \left(2 \log_2 x + 5 \right) \][/tex]
Since the derivative of [tex]\( \log_2 x \)[/tex] is [tex]\(\frac{1}{x \ln 2}\)[/tex], we get:
[tex]\[ k'(x) = \frac{2}{x \ln 2} \][/tex]
Setting [tex]\( k'(x) = 0 \)[/tex] to find critical points:
[tex]\[ \frac{2}{x \ln 2} = 0 \][/tex]
This equation has no solutions because the derivative [tex]\(\frac{2}{x \ln 2}\)[/tex] is never zero for [tex]\( x > 0 \)[/tex].
3. Equation of a Line:
Consider a generic line [tex]\( y = mx + b \)[/tex].
4. Form the System of Equations:
Set [tex]\( k(x) \)[/tex] equal to the line:
[tex]\[ 2 \log_2 x + 5 = mx + b \][/tex]
5. Solve for x:
To find solutions for [tex]\( x \)[/tex], rewrite the logarithmic equation:
[tex]\[ 2 \log_2 x + 5 = mx + b \][/tex]
Converting [tex]\(\log_2 x\)[/tex] to natural logarithms ([tex]\(\log_2 x = \frac{\ln x}{\ln 2}\)[/tex]):
[tex]\[ 2 \frac{\ln x}{\ln 2} + 5 = mx + b \][/tex]
6. Analyzing the Solutions:
Rewriting this, we get:
[tex]\[ 2 \ln x + 5 \ln 2 = mx (\ln 2) + b \ln 2 \][/tex]
[tex]\[ 2 \ln x + 5 \ln 2 = mx \ln 2 + b \ln 2 \][/tex]
To find the points of intersection, solve this equation for [tex]\( x \)[/tex]:
[tex]\[ 2 \ln x + 5 \ln 2 = mx \ln 2 + b \ln 2 \][/tex]
This has solutions for [tex]\( x \)[/tex] in terms of the Lambert W function. The solutions are:
[tex]\[ x = -\frac{2 \text{LambertW}(-\sqrt{2} m \sqrt{e^{b \ln 2}} \ln 2 / 16)}{m \ln 2} \text{ and } x = -\frac{2 \text{LambertW}(\sqrt{2} m \sqrt{e^{b \ln 2}} \ln 2 / 16)}{m \ln 2} \][/tex]
These results indicate that under certain conditions for [tex]\( m \)[/tex] and [tex]\( b \)[/tex], the line [tex]\( y = mx + b \)[/tex] can intersect the function [tex]\( k(x) = 2 \log_2 x + 5 \)[/tex] at three points, but it depends heavily on the specific values of [tex]\( m \)[/tex] and [tex]\( b \)[/tex]. Generally, you would select [tex]\( m \)[/tex] and [tex]\( b \)[/tex] to ensure that the logarithmic expression results in real values for the solutions involving the Lambert W function.
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