 We have seen previously that if we're given the formula of a logarithmic function that we can produce its graph But what if you have to go the other way around? What if we have the graph of the function? We want to come up with a formula Let's say that for example We need to find the base a so that the graph f of x which is equal to log base a of x contains the point two comma three So we know that we have a just a standard logarithmic function no transformation whatsoever But it contains the point two two comma three here for which We're aware that the base does affect how steep the logarithm is going to be And so let's say that we have that point there. So we have the two two is the x coordinate Three is the y coordinates. We have something like this So we have this point two comma three Since our logarithm hasn't been transformed whatsoever. We do know it also will contain just the standard x intercept one Like so and so if we try to think about what this logarithm would look like It's it's vertical asymptote still be the y-axis. We would see graphs something like this So this is f. What if we know about this? What if we know this about the function? Like we could see this from the picture Well, we can use this point to figure out what the base of the function is Because after all if we were to look at f of two Well because of the point we know that f of two is going to equal three But because of the function formula, we know f of two is going to equal log base a of two And so we have this equation log base a of two is equal to three Now we want to solve for a but like how do you how do you solve for a right here? um We can't really move the two over because of the logarithm What we want to do is actually move the log base a Over that's something we know how to do we can switch it from logarithmic form to exponential form This would look like two is equal to a cubed All right, so we switched it from logarithmic form to exponential form But in this situation the unknown is actually the base the power is what we know So we can get rid of the three we can move the three back to the other side for which We take the cube root of both sides and we see that here the base a Should equal the cube root of two for which we can approximate that but we're just going to keep an exact form right here So this logarithm the logarithm the standard logarithm that goes to the point two comma three must have been base uh cube root of two Now of course things can get a little bit more complicated like look at this picture We see right here. We see that we have a logarithm which goes through the point One zero it looks like the standard uh x intercept It also goes to the point two one But look at the look at the vertical asymptote the vertical asymptote is at x equals three How does that affect things? Yikes Well in order to in order to decide what's going to go on here What I wanted to do is first convince you that the most general form of a logarithm is this formula Let me see right here f of x equals log base a of x minus h over b That's the that's the worst case scenario when it comes to a logarithm And how do I know that well in a previous video? We saw that the most general exponential function we could graph would look like y equals c a to the x plus k Right where the parameter c which gives you the vertical stretch the base a and the vertical shift k That's the only thing you can do with a with a x-minitial function All ever all other transformations can be baked into this cake Well, let's solve for x in this situation right here. Let's let's compute the inverse function So actually if I were to swap x and y you would get x equals c a y Plus k and so if you solve for y in this situation right the first thing to do is subtract k from both sides You're going to get you're going to get x minus k is equal to c a to the y Then you're going to divide both sides by c So you're going to get a to the y is equal to x minus k Over c and then you switch it to logarithmic form you get y equals the log base a Of x minus k over c like so And so you can see what happened is now we get something like this that although the labels h and b are different from k and c I switched these to be h and b because we use h for horizontal shifting and we use b for You know vertical horizontal stretching and compressing things like that But this right here if this is the most general exponential function Then the most general logarithm would look something like this or again what we see right here That is to graph a exponential function. We only need vertical transformations We just need a stretch and a shift vertically if we know the base All right because of that since Since vertical transformations turn into horizontal transformations when you go to the inverse function the most general logarithm can be expressed using just just horizontal transformations So how do these these horizontal transformations affect the graph? Well the standard logarithm should have It should have as its vertical acetote the x the y-axis excuse me But this one right here has gone one two three steps to the right And so imagine that if you have your graph and you shifted everything to the right That's the only thing that's going to change this vertical acetote because the horizontal compressions and stretches won't affect the y-axis And the base the base doesn't affect the y-axis as well So the only thing that can move the vertical acetote will be this shift right here And so I want you to I want you to be aware of that h H right here is going to correspond to the location The location of the vertical acetote Which in this in this graph here We can see very quickly that because the vertical acetote is located at x equals three We see that h is going to equal three in this situation All right now. It's going to be a positive three That's important to remember if we had like over here It was like negative two then we plug in a negative two because you get a negative negative two Which give you a plus two. It's a shift to the left. All right, so the location of the Ascentote tells you the shift we did the exact same thing for exponentials after all As we move the horizontal acetote up and down that would change the vertical shift It was the same number. We see the the inverse relationship going on here The location of the vertical acetote gives you the horizontal shift Well, how did we find the location or how do we figure out how much of a vertical stretch or compression happened to? How how how did that affect an exponential? We looked at the y-intercept and the distance it was from the horizontal acetote We're going to do the same thing, but we're going to take the x-intercept We're going to take the x-intercept and measure the distance from the x-intercept to the vertical acetote Take the x-intercept and minus the vertical acetote In this situation here. So we're going to take one So this this number right here is going to be this b value. So to find b We're going to take the difference The difference between the x-intercept And the Vertical acetote so we see in this situation b is equal to 1 minus 3 which gives us a negative 2 All right, so let's kind of summarize what we have so far So what we know about our function is the following f of x is equal to the log base a we don't know what that is yet But we're going to get x minus 3 over negative 2 We have a negative 2 in the denominator now It's a little bit disconcerting to have a negative inside of a logarithm It's not actually wrong, but I'm actually going to rewrite this as the log base a Of 3 minus x over 2 that makes me feel a little bit happier here Now we're in a situation where we have to figure out what is the base a and that's where we're going to use the point We haven't used yet. We know the point 2 comma 1 is on the graph I'm going to plug that into the equation to solve for the the yet still missing base So we know that f of 2 Is going to equal 1 but we also know it's going to equal the log base a of 3 minus x well 3 minus 2 actually Over 2 Like so for which we can simplify this we get the log base a of 1 half This is supposed to equal 1 and just like we did on the previous example. We're going to move the base a to the other side And this then gives us 1 half Is equal to a to the first we're just as equal to a now if if we had some other number than one We would then have to uh, we'd have to you know, take a square root or cube root or whatever And so this is this is a great point for us that when it comes to logarithms We like the x-intercept and we like the point Where the y-coordinate is equal to 1 we saw the same thing with exponential functions We like the y-coordinate and we like the place where the x-coordinate was equal to 1 That is we like the y-intercept and when the x when the x-coordinate was 1 for logarithms It's the opposite We want the x-coordinate and when y equals 1 but if we could do with any point here because again We could take we could take like a second like square root or cube root like we did on the previous example So since the base is equal to 1 half We now have our function f of x equals the log base 1 half as much as you might not want a fraction right here It is perfectly acceptable. The base of a logarithm can be any positive number other than 1 So we get the log base 1 half Of 3 minus x over 2 and this function right here will then produce the graph that we started off with right here