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CCSS.Math: ,

so we have an interesting equation here and let's see if we can solve for K and we're going to assume that M is greater than zero like always pause the video try it out on your own and then then I will do it with you all right let's let's work on this a little bit so you could imagine that the key to this is to simplify it using our knowledge of exponent properties and there's a couple of ways to think about it first we could look at this rational expression here M to the 7/9 power divided by M to the 1/3 power and the key realization here is that if I have if I have X to the a over X to the B then this is going to be equal to X to the a minus B power it actually comes straight out of the notion that X to the a over X to the B X to the a over X to the B is the same thing as X to the a times 1 over X to the B which is the same thing as X to the a times 1 over X to the B that's the same thing as X to the negative B which is going to be the same thing as if I have a base to 1 exponent times the same base to another exponent that's the same thing as that base to the sum of the exponents a plus negative B which is just going to be a minus B so we got to the same place so we can rewrite this as so we can rewrite this part as being equal to M to the 7/9 power minus 1/3 power is equal to is equal to M to the K over 9 and I think you see where this is going what is seven ninths minus 1/3 well one-third is the same thing we've won have a common denominator one-third is the same thing as three nights so I can rewrite this as three nights so seven ninths minus three nine is going to be four ninths so this is the same thing as M to the M to the 4/9 power is going to be equal to M to the K nights power so four ninths must be the same thing as k nights so we can say 4/9 is equal to K nights four over nine is equal to K over nine which tells us that K must be equal to four and we're all done