Log Function Transformation - Find values of 'a' and 'k' given two transformed points

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DaRafster

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Question: The graph of [MATH]f(x) = log_2x[/MATH] has been transformed to [MATH]g(x) = alog_2x+k[/MATH]. The transformed image passed through the points (1/4, -9) and (16, -6). Determine the values of a and k.

I'm not entirely sure how to determine the values of a and k. My initial observation were that only the y values change and not the x values. So, I proceeded to find the parent points:

Transformed Point: (1/4, -9)
[MATH]y=log_2(1/4)[/MATH][MATH]2^y = 1/4[/MATH][MATH]ylog_2 = log(1/4)[/MATH][MATH]y=log(1/4)/log2[/MATH][MATH]y=-2[/MATH]Parent Point: (1/4, -2)

I then did the same thing for the other transformed point and got (16, 4) as the parent point.

As to what I do next, I have no idea. I'm not even necessarily sure if you're required to find the parent points, I just solved for them because it was the only thing I could think of.

What happens next after I find the parent points? Or what should I have done differently?
 
skeeter said:
[MATH]-9=a\log_2(1/4) + k = -2a + k[/MATH]
[MATH]-6 = a\log_2(16) +k = 4a +k [/MATH]
two equations, two unknowns … solve the system
Click to expand...

Thanks for the reply.

My work:
[MATH]-9=-2a+k[/MATH][MATH](-9-k)/-2 = a[/MATH]
Is this what you mean by solve the system? Sorry, I still find it a bit unclear. I'm still unaware how I'm supposed to solve for a or k given the equation with two unknowns.
 
DaRafster said:
Thanks for the reply.

My work:
[MATH]-9=-2a+k[/MATH][MATH](-9-k)/-2 = a[/MATH]
Is this what you mean by solve the system? Sorry, I still find it a bit unclear. I'm still unaware how I'm supposed to solve for a or k given the equation with two unknowns.
Click to expand...
[math]-9 = -2a + k[/math]
[math]-6 = 4a + k[/math]
Both of these equations are true at the same time so they both have the same values of a and k. You solved the top equation for [math]a = \dfrac{-9 - k}{-2} = \dfrac{k + 9}{2}[/math] so put that into the second equation and you have an equation just in k.

[math]-6 = 4 \left ( \dfrac{k + 9}{2} \right ) + k[/math]
-Dan
 
topsquark said:
[math]-9 = -2a + k[/math]
[math]-6 = 4a + k[/math]
Both of these equations are true at the same time so they both have the same values of a and k. You solved the top equation for [math]a = \dfrac{-9 - k}{-2} = \dfrac{k + 9}{2}[/math] so put that into the second equation and you have an equation just in k.

[math]-6 = 4 \left ( \dfrac{k + 9}{2} \right ) + k[/math]
-Dan
Click to expand...
Oh, that makes sense! Thanks for the clarification!
 
[MATH]-6 = 4a+k[/MATH][MATH]-9 = -2a+k[/MATH]
subtract the second equation from the first ...

[MATH]3 = 6a \implies a = \frac{1}{2}[/MATH]
substitute [MATH]\frac{1}{2}[/MATH] for [MATH]a[/MATH] in either equation to find [MATH]k[/MATH]
 
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