Understanding the Horizontal Shift in Logarithmic Functions

[zekea]

If we have log 3 (x+9) + 2 = y. It states that we have a HT 9 units left and 2 units up. But if we manipulate that too

log 3 (x+9) = y - 2

And convert to Exp form. This gives me

3^(y-2) = x + 9 or 3^(y-2) - 9 = x

This looks to me more like a HT 2 units right and a vt 9 units down. I don't understand why x+9 is a HT when changing to exp form it looks like your K value (Vertical Translation)

 

[Jameson]

If we have log 3 (x+9) + 2 = y. It states that we have a HT 9 units left and 2 units up. But if we manipulate that too

log 3 (x+9) = y - 2

And convert to Exp form. This gives me

3^(y-2) = x + 9 or 3^(y-2) - 9 = x

This looks to me more like a HT 2 units right and a vt 9 units down. I don't understand why x+9 is a HT when changing to exp form it looks like your K value (Vertical Translation)

Hi zekea!

Lets take a look at this graph.

[desmos]y=log_3 (x+9) + 2 [/desmos]

All of the math you did with transforming the starting equation is correct, but maybe I can help you see why we have a horizontal shift.

Let's start with just $y=\log(x)$. You can plot that in the interactive graph in this post if you like. An easy point on this graph to get is $\log(1)$. This is 0 for any base, meaning $\log_2(1)=\log_3(1)=\log_{\pi}(1)=0$. So on this graph we have the point (1,0). Now what about $\log(x+9)$? When $x+9=1$ or $x=-8$ we get the $\log(x+(-8))=\log(1)=0$. How does this compare to our starting point of $\log(x)$? We get the same y-value when we shift the x-value 9 units to the left.

If you play around with values on the graph you'll see this kind of behavior. Does that help at all? :)