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### Course: Algebra (all content)>Unit 11

Lesson 30: Graphs of logarithmic functions (Algebra 2 level)

# Graphs of logarithmic functions

The graph of y=log base 2 of x looks like a curve that increases at an ever-decreasing rate as x gets larger. It becomes very negative as x approaches 0 from the right. The graph of y=-log base 2 of x is the same as the first graph, but flipped over the x-axis. The graph of y=-log base 2 of (x+2) is the same as the first graph, but shifted to the left by 2. Created by Sal Khan.

## Want to join the conversation?

• At , why is it log base 2 (x+2) and not (x-2)? The blue graph is shifted 2 points to the left and not the right.
• In terms of the x - h example above, I've been finding it helpful to think that "the graph must compensate for h".

E.g. when you see log_2(x-2), you have "lost" 2 units, and must compensate by moving the graph 2 units to the right. Likewise when you see log_2(x+2), you have "gained" 2 units, and must compensate by moving the graph 2 units to the left.
• even after watching all the videos related to logarithmic functions I still can't comprehend let alone complete the practice session! Can you post another video that thoroughly explains how to find which equation matches the graph.
• You can always just plot points for each graph and see which one fits. It will take a bit longer though.
• How do you actually do the math for a negative logarithm?
• Do you mean the negative of a logarithm, or the logarithm of a negative number?
The logs of negative numbers (and you really need to do these with the natural log, it is more difficult to use any other base) follows this pattern.
Let k > 0
ln (−k) = ln (k) + π 𝑖
For other bases the pattern is:
logₐ(−k) = logₐ(k) + logₐ(e)*π 𝑖

If you mean the negative of a logarithm, such as
y = − log x, then you just change the sign of y as you would with any other negation.
• What's up with that red guy who appears sometimes above a video and then disappears?
• to see if you are actually watching the videos! Hahaha
• At Sal states that 2 to the 0th power is 1, why? Shouldn't it be 0 because anything times 0 = 0? Plus 2 times itself 0 times...I'm confused. :l
(1 vote)
• The easiest way to explore is this:
Anything divided by itself is 1. [Except 0, 0/0 is undefined]
So (n^a)/(n^a)=1

What happens when divide bases to powers? We subtract them.
( n ^ a ) / ( n ^ a ) = n ^ ( a - a ) .
a-a=0.
Therefore
n ^ 0 = n ^ ( a - a ) = ( n ^ a ) / ( n ^ a ) = 1
n ^ 0 = 1.

However, since 0/0 is undefined, 0^0 is also undefined. So it is a MOSTLY true statement. One exception, however, isn't bad when you realize losing that exception breaks mathematics.
• For the 4th option, where y = - log2(x+2) , if I put y= -2 , I get x=14, which doesn't match the curve he's drawn?
(1 vote)
• See if this helps:
Like Sal mentioned, ask yourself, "What power (i.e y) of the base (i.e 2) will give x"?
With y=-2, we have to find 2^-2. We can write this in the form 1/(2^-2), which is 1/4.
Also, check out the earlier video on rules of exponents.
• This is easier to understand, but is there a faster way of calculating this?
• when graphing log 2 X, I get confused. you keep saying that X is equal to 8, but isn't Y technically equal to 8? X is the exponent, not the total? please elaborate.
(1 vote)
• y=log2x is the same thing as 2^y=x so if x=8 we can subsitute it in.
2^y=8
we know that 2^3=8 so y=3