Vertical shifts

Question

graph y=logx, y=log(10x), and y=log(100x). how do the graphs compare? use a property of logs to show that the graphs are vertical shifts of one another. 

Answer 1

graph y=logx, y=log(10x), and y=log(100x). how do the graphs compare? use a property of logs to show that the graphs are vertical shifts of one another.

Eric's explanation is confusing because of the typos in it.

So let me try:

f(x) = log(x)

g(x) = log(10x) = log(10^1) + log(x) = log(x) + 1 = f(x) + 1

h(x) = log(100x) = log(10^2) + log(x) = log(x) + 2 = f(x) + 2

So g is f shifted up 1 and h is f shifted up 2.

Eric also explained dilation ("Honey, I shrunk the kids!"), but that's not needed for this problem.

Here’s a GeoGebra graph of the three functions:
http://www.wyzant.com/resources/files/265623/logs_translated

Answer 2

There are a few different ideas you need to know here.

 

GRAPHING FUNCTION CONCEPTS

When graphing, y = f(x)...

 

Stretch / Compression

Replacing x with (ax) is a horizontal stretch/compression. y = f(ax)

Multiplying to get y = a*f(x) is a vertical stretch/compression.

 

You should also remember

For vertical moves, if a>1, the graph is stretching out vertically.

For horizontal moves, if a>1, the graph is compressing in horizontally. 

 

Shifts

Replacing x with (x-b) is a horizontal shift. Moves entire graph left or right

y = f(x) + b is a vertical shift. Moves entire graph up or down

 

 

In this case function y= f(x) is y=logx.

PROPERTY OF LOGS

log_x(a*b) = log_x(a)+log_x(b)

 

When a base x is not shown, it is commonly assumed base x is 10.

 

log₁₀(10^a) = a

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OK so knowing all this, now we can start the problem... Wow.

 

y = logx      AND     y = log10x

 

y = log(10x), we are replacing x with 10x. x>1 Therefore, this is a horizontal compression.

 

y = log10x = log10 + logx

y = 1+logx

y = logx + 1   Therefore, this is vertical shift of y = logx. The graph moves up 1.

 

I know this is a lot of information, but if you take the time to understand all these ideas, you will likely be ahead of your class!