 We can solve exponential equations in two ways. If we're fantastically lucky and have two exponential expressions with the same base equal to each other, so for example, if a to power x equals a to power y, then we know that x is equal to y, and conversely. In a kind and gentle universe, we'd always have exponential expressions with the same base. We don't live in that universe, so we more often rely on the following. If a to power x equals y, then we can hit both sides with a log, log of a to power x equals log of y. So, let's suppose we're fantastically lucky and get an exponential expression where the bases are the same. Since the bases are the same, if a to power x equals a to power y, then x equals y. So our two exponents, x minus 5 and 8 minus 3x, have to be equal, and we can solve getting our solution. And this is probably about the only time we'll ever be so fortunate as to have two exponential expressions with the same base. So let's try to solve this. Since the bases are equal, we can write the exponents equal. Oh wait, that's not right. I was just told the other day that 150,000 is almost nothing, so 5 is 25. But for those of us who like to live in the real universe, 5 is not equal to 25, and we can't equate the exponents. Now, what we could do is, since 25 is 5 to the second, we can rewrite, we can apply the rules of exponents, and now our bases are equal, and so we can equate the exponents and solve. And again, it's worth taking a longer, appreciative look at this problem because this is likely the only time we'll ever be able to change one base into the other. More often we'll probably have to do something like this, and so here it's useful to remember we can hit both sides with a log. Now, we do need to remember the rules of logs. Remember the log of a power allows us to move the exponent out front, and log of 7 is just a real number so we can distribute this product. And let's isolate our x terms, and let's go ahead and simplify the left-hand side here. 8 log 7 minus log 50 can simplify, and solving for x, and if we want to know what this value is, we'll use a $10 scientific calculator. One useful thing is that because we can handle products and quotients with logs, as long as our expression doesn't have any additions or subtractions, we can hit both sides with a log. So here, let's hit both sides with a log. Now, on the left-hand side we have the log of a product, and our rules of logs say the log of a product is the sum of the logs. We have the log of a power, so we can rewrite that. We have the log of a quotient, so we can rewrite that. The log of 1 is 0, and so our left-hand side simplifies. Again, let's isolate our x, and we have a difference of logs which can be rewritten as a quotient, and we can solve for x.