 If y is equal to a to the power x, then x is the log-to-base a of y. And so the graph of y equals log ax can be found by reflecting the graph of y equals a to the x across the line y equals x. We'll focus on the cases where a is greater than 1. Now, this means a couple of important things. Since the graph of y equals a to the power x has y greater than 0 for all x, then the graph of y equals log a of x will have x greater than 0 for all y. And since the graph of y equals a to the power x has asymptote of y equals 0, then the graph of y equals log a of x has asymptote of x equal to 0. So, for example, the graph shown is a transformation of log-to-base 2 of x using only horizontal translations and vertical stretches. If the graph passes through the point 7, 4, describe the transformations and write the equation. Now, the useful thing to remember here is that the graph of y equals log x will have a vertical asymptote at x equals 0. Now, this graph appears to have a vertical asymptote at x equals 3, which suggests a horizontal translation of 3 units to the right. And so we'll make that our first transformation. The graph can be produced from y equals log 2 of x by shifting 3 units to the right, producing the graph of y equals log-to-base 2 of x minus 3. Now, since we're told the graph has a vertical stretch, we know it will have the form y equals some constant times log 2 x minus 3. And since it passes through the point 7, 4, we know that x equals 7, y equals 4, makes this equation true. And we can now solve this equation for c, which tells us that the graph is stretched vertically by a factor of 2 to produce the graph of y equals 2 log 2 x minus 3. Or let's take another graph and let's see if we can find a different set of transformations to produce the same graph. This time, our graph has just horizontal and vertical translations. So again, a good starting point is to remember that the graph of y equals log of x has an asymptote at x equal to 0. This graph appears to have an asymptote at x equals negative 2, which suggests a horizontal shift of 2 units to the left. And so the graph of y equals log x is translated to units to the left to produce the graph of y equals log of x plus 2. Now, if there's a vertical translation, the graph will be that of y equals log x plus 2 plus some constant k. Since the graph goes through 8, 0, then x equals 8, y equals 0 is a solution, which means we can substitute these values in and solve for k. And we find k equals negative 1, and so the graph is shifted downward by 1 unit to obtain the graph of y equals log x plus 2 minus 1. To find another transformation, remember we can combine logarithmic expressions. So one useful thing is that log of 10 is equal to 1, and so this minus 1 is the same as minus log 10. But now I have a difference of logs, and a difference of logs can be rewritten as a quotient. And so this is y equals log of x plus 2 divided by 10. And so this suggests a different set of transformations. We can stretch the graph of y equals log x horizontally by a factor of 10 to produce the graph of log of x divided by 10. Then shift the graph horizontally by two units to the left to produce the graph of y equals log of x plus 2 divided by 10.