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

DaRafster

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Jan 25, 2020
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Question: The graph of \(\displaystyle f(x) = log_2x\) has been transformed to \(\displaystyle g(x) = alog_2x+k\). 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)
\(\displaystyle y=log_2(1/4)\)
\(\displaystyle 2^y = 1/4\)
\(\displaystyle ylog_2 = log(1/4)\)
\(\displaystyle y=log(1/4)/log2\)
\(\displaystyle y=-2\)
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

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\(\displaystyle -9=a\log_2(1/4) + k = -2a + k\)

\(\displaystyle -6 = a\log_2(16) +k = 4a +k \)

two equations, two unknowns … solve the system
 

DaRafster

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\(\displaystyle -9=a\log_2(1/4) + k = -2a + k\)

\(\displaystyle -6 = a\log_2(16) +k = 4a +k \)

two equations, two unknowns … solve the system
Thanks for the reply.

My work:
\(\displaystyle -9=-2a+k\)
\(\displaystyle (-9-k)/-2 = a\)

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.
 

topsquark

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Thanks for the reply.

My work:
\(\displaystyle -9=-2a+k\)
\(\displaystyle (-9-k)/-2 = a\)

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.
\(\displaystyle -9 = -2a + k\)

\(\displaystyle -6 = 4a + k\)

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 \(\displaystyle a = \dfrac{-9 - k}{-2} = \dfrac{k + 9}{2}\) so put that into the second equation and you have an equation just in k.

\(\displaystyle -6 = 4 \left ( \dfrac{k + 9}{2} \right ) + k\)

-Dan
 

DaRafster

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\(\displaystyle -9 = -2a + k\)

\(\displaystyle -6 = 4a + k\)

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 \(\displaystyle a = \dfrac{-9 - k}{-2} = \dfrac{k + 9}{2}\) so put that into the second equation and you have an equation just in k.

\(\displaystyle -6 = 4 \left ( \dfrac{k + 9}{2} \right ) + k\)

-Dan
Oh, that makes sense! Thanks for the clarification!
 

skeeter

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Dec 15, 2005
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\(\displaystyle -6 = 4a+k\)
\(\displaystyle -9 = -2a+k\)

subtract the second equation from the first ...

\(\displaystyle 3 = 6a \implies a = \frac{1}{2}\)

substitute \(\displaystyle \frac{1}{2}\) for \(\displaystyle a\) in either equation to find \(\displaystyle k\)
 

Jomo

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Dec 30, 2014
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You do not need logs to solve 2y=1/4for y. y=-2
 
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