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

#### DaRafster

##### New member
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

##### Elite Member
$$\displaystyle -9=a\log_2(1/4) + k = -2a + k$$

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

two equations, two unknowns … solve the system

• HallsofIvy and topsquark

#### DaRafster

##### New member
$$\displaystyle -9=a\log_2(1/4) + k = -2a + k$$

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

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

##### Senior Member
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

##### New member
$$\displaystyle -9 = -2a + k$$

$$\displaystyle -6 = 4a + k$$

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

##### Elite Member
$$\displaystyle -6 = 4a+k$$
$$\displaystyle -9 = -2a+k$$

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

##### Elite Member
You do not need logs to solve 2y=1/4for y. y=-2