 So the next step after introducing a logarithm is to consider equations that involve logarithms. So let's jump right in. Let's say we want to solve log to base 10 of 3x plus 7 equals 2. So definitions are the whole of mathematics. We know that from this equation we can rewrite it in exponential form 10 to power 2 equals 3x plus 7. And this gives us a fairly nice equation and we can get our solution. As with other types of equations, we'll want to check our solution. And again, it's useful not to simplify our arithmetic expression. So we'll check by dropping this value into our original equation. And remember the log is the exponent. So the log to base 10 of 10 to the second is in fact equal to 2. So our statement is true. And so x equals 10 to the second minus 7 over 3 is a solution. How about this equation? The key thing to remember here is that we know how to rewrite log equals but this isn't in the form log equals. So we should do some algebra. In this case our rules of logarithms allow us to combine logs. This is the log of something minus the log of something else. And so this is reminiscent of the rule for the log of a quotient. The log of a quotient is the difference of the logs. So we can rewrite this as a single logarithmic expression. Now our equation is in the form log equals something. So we can use our definition of logs to rewrite this 2 to power 3 equals x plus 10 over x plus 1. And that gives us a nice rational equation to solve. So we'll solve that equation. And again we should check in the original equation. We'll substitute in what we think our solution is. Now one problem is that we can't calculate logs directly. So we'll have to use our rules of logs. But before we do that we have to keep in mind one very important idea. Log to base a of c is only defined for positive values of c. Which means that whatever we're taking the log of here has to be positive. If it isn't then our solution automatically fails and is not actually a solution. Fortunately we are taking the logs of positive numbers so we can combine them using the rules of logs. This is the difference of logs. Well that's the log of a quotient which we can invert and multiply. Do a little arithmetic. And rather than trying to find the log to base 2 of 8, the thing to remember is the log is the exponent 2 to power 3 is 8. So this last statement is true and our solution checks. So here we have another logarithmic equation and we want to use our rules of logarithms to get the equation into the log equals something form. So we have the sum of logs so we can rewrite it as the log of a product. Then we can use the definition of logs to rewrite this where remember if the base isn't stated we can assume that our base is 10. And now this gives us a nice quadratic equation. So we have product equal to 10 which means that our individual factors have to be well we can't actually do anything with product equal to 10 so we expand then solve. So we should always check our solutions. If x is 4.099 then substituting this into our logarithmic equation gives us now in general finding actual logarithmic values is very difficult. So we might not be able to check to see if this is a true statement. What we would at least like to make sure of is that we are not trying to take the log of a negative number because remember log is only defined for positive numbers. So both of the arguments here are positive so provided we've done the other algebra correctly x equals 4.099 does seem to be a solution. On the other hand if x equals negative 6.099 then substituting these values into our original equation gives us and here we have a problem the first argument and for that matter the second are both negative and we can't take the log of a negative number so x equals negative 6.099 can't be a solution. It's an extraneous solution and so the only solution is x approximately 4.099.