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Introduction to Logarithms  Concept
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
To solve rate word problems, knowledge of solving systems of equations is necessary. Rate word problems include problems dealing with rates, distances, time and wind or water current. Other types of word problems using systems of equations include money word problems and age word problems.
So for this episode what I want to do is take a look at the graph of f of x is equal to 2x. Okay, so I have my exponential function and I have a rough sketch of the graph.
So looking at this we know that this graph is a function because for every x there's only one y. We also know that it is 1:1 because for every y there is one x. So because there it's 1:1, we know that the this graph, this function will have an inverse. What I want to do is find out what that inverse is. So if you remember back to our inverses, what we do is f of x is the same thing as y, so that y is equal to 2x and then to find the inverse we just switch our x and our y. So we end up with x is equal to 2 to the y. Okay.
Our last step in finding the inverse was to solve this for y. The problem is, is that using the methods that we know there is no way to solve this for y. Okay? So what has been done is we have introduced a new function altogether. A logarithm function, okay? And how that works is this is the log base 2 of x is equal to y. These two statements are exactly the same, okay? This is called exponential form and this one over here is logarithmic form. And I'm a horrible speller, do hopefully I got that right.
Key thing to remember, okay, and it's kind of hard to get used to this new log based this is a little subscript, sort of a new form but basically it's the exact same thing as this. What I do to remember this, okay, is two things. First is number 2 is called the base when you're dealing with exponential forms. The base of our exponential form becomes the base of our log, okay? So the base stays the base. The other way I tend to remember this is that if I'm putting something away from log formula. exponential form and we're going to spend some time going back and forth between these two, is that the 2 just sort of comes up and under the y and bumps the y up. So the 2 will come over up the y up and the log falls out, okay? So those are sort of the 2 ways you can remember it. The base stays the base and then then the 2 so that the base sort of comes up and under bumps the other side up, log falls away. Okay? Little bit of lingo to go along with this just so you know how to talk about this form. This is this is log base 2 of x, and obviously the 2 in this case is a completely random number. It could be a 3, it could be a 7, it could be 124 but basically, you say log, base number of whatever your inside function is in order to talk about this, okay.
So starting with our exponential graph, we found the inverse. We couldn't solve for y so we invented this new way of writing it, okay? The 2 different ways of writing it and just thought of one other thing I want to talk about this [IB] number over here, is log base a of xy is basically [IB] of x so this side is the power to which you raise a to get x. Okay? Just in sense form I know that often [IB] hard to distinguish especially when I subtract like this but basically this is the same the sense is the same thing as this. We're trying to figure out a to what power will give us x. Okay?
So logarithms. Basically it's the inverse of an exponential we cancel out for it so we had to invent this new function in order to deal with it. But hopefully as you deal with it you'll sort of understand how the whole logarithm thing works and it's not something too scary.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
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Introduction to Logarithms
Problem 1 9,785 viewsPut into logarithmic form.
25 = 5²3^{4} = 1 81 
Introduction to Logarithms
Problem 2 7,561 viewsPut into exponential form:
log_{3}9 = 2log_{⅕}125 = 3 
Introduction to Logarithms
Problem 3 1,429 views 
Introduction to Logarithms
Problem 4 1,365 views 
Introduction to Logarithms
Problem 5 1,257 views
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