Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Exponential equations if B to the M equal B to the N then M equal N. Logarithmic equations If log base B of M equal log base B of N then M equal N Let's look at some examples. 3 to the two X + 1 equal 81. Well 81 we know is really the same thing as 3 to the 4th power. So if I can get the basis to be the same, then I know the exponents have to be equal. And then we can solve for X. Subtract the one divide by two X equal 3 halves eight to the 1 -, X equal 4 to the X + 2. Eight and four both have a common base of 2, so 8 is 2 ^3 and four is 2 ^2. Now powers to powers mean we multiply. So we're going to get 2 to the 3 -, 3 X equaling 2 to the two X + 4. Once I can get those bases the same, we set the exponents equal. If I take all the X's to one side, I get 5X equaling -1 X equal, negative 1/5. For this next one, 9 is really 3 ^2 and one over the cube root of 3. Well, let's start with the cube root is really to the one third power. Then powers to powers are multiplied. And how do I get rid of the one over? Well, we put a negative in the exponents so that we take the reciprocal. Once the bases are the same, we set the exponents equal and we're going to get X equal -1 sixth. Now sometimes we get problems that don't have the same bases and we can't make them the same. What we can do is we can take the logarithm of each side in base 10 or we could use base E. Now if we took that X which is in the exponent down in front using the property rule then we could get the X by itself by dividing each side by log 5. So log 17 divided by log 5. Another way of doing the same problem is to think about the exponent equals log base 5 of 17, and then we could take that into the change of base formula, which would say log of 17 over log of five. The next one, we need to get the base with the exponent by itself first. So we're going to add 2 to each side to get rid of the E That's really the natural log. If we put it in logarithmic form, so the exponent equals the natural log or the log base E of 10478. To solve for X, we're going to add 3 to the other side. And it doesn't matter which order we add in, because we have the commutative property and then we're going to divide by 5. If you wanted a decimal approximation, you'd just put that into your calculator. The next example we're going to solve by factoring what times what gives us E to the 4X? Well, E to the two X * e to the two X because if we multiply we add the exponents what are two numbers that multiply to give us -24 and add to give us 5 positive 8 and -3. So E to the two X equal -8 and E to the 2X has to equal 3. To get rid of this E with two X equals the natural log of -8. Well, we can never have the natural log or log of any negative number, so this part is not possible. If we look at this other part, 2 X equals the natural log of three. That's OK. So X equals the natural log of 3 / 2. Still a few more examples. If we start with logarithmic form and we want to solve for the X, we're going to have the base to the exponent. Remember the log base 5 of X is really an exponent. That whole thing together equals X. SO5 cubed is X or 1:25 for the next one we're going to take, and we're going to put the six divide each side. So lane of two X equal 5, E to the fifth equal 2X, or X = e to the 5 / 2. Now there's more than one way to do this. One, we could have also taken that six up into the power. Then when we changed this, we would have had two X to the 6th power equaling E to the 30th. How do we get rid of a 6th power? We take it to the 6th root. So the 6th root of E to the 30th 6 divides into 35 times and then if we would have divided by two, we would have had E to the fifth over 2. Same answer, two different ways to get there. This next one, if I take the three up into the exponent, then I get a log equaling a log and the spaces are the same. So X ^3 equal 125, X equals the cube root of 125 which is just five. If we have a log equaling a log plus a number, we have to get all the logs on one side first. So we're going to subtract that log base three of nine, and let it equal 2. Now if we have a coefficient, we know that it has to go up as a power, and if we're subtracting, we know that that's actually going to go in the denominator and rewrite it as a single log. So with that, then we're going to be able to change it into exponential form. An exponential form says that we're going to have 3 ^2 equaling X + 4 ^2 / 9. Well 3 ^2 is 9 and to get rid of the denominator 9 * 9 is 81 = X + 4 quantity squared square root each side. So positive or -9 equal X + 4 subtract 4 so we get X equal 5 or X equal -13. Now when we look at the back into the original equation, if I put in X equal 5, do I always get a positive logarithm? And the answer is yes. But if I put in X equal -13 -13 + 4 is -9 and we can't ever have a logarithm of a negative, so -13 is not going to work if we have a log minus a log equaling a log. We can actually rewrite this with the +1 in the numerator and the -1 in the denominator. And whenever a log equals a log, we can drop the logs and just solve for the inner peace. So X - 1 / X + 3 equal 1 / X cross multiply. So X ^2 -, X equal X + 3 Need to get everything to one side. So X ^2 - 2 X -3 = 0 factor that X - 3 * X + 1 X equal 3 X equal -1. Now we need to check that back in the original. If we check that back in the original, X equal three 3 - 1 is +3 + 3 is positive and 1 / 3 is positive. So X equal 3 is going to work. If I put -1 negative 1 -, 1 is -2 we don't know the log of a negative number so that one will not work. Thank you and have a wonderful day.