Solving exponent equation using exponent properties

So I have an interesting equation here. It says ( V^{-65} ) times the fifth root of ( V ) is equal to ( V^{K} ) for ( V ) being greater than or equal to zero. What I want to do is try to figure out what ( K ) needs to be. So what is ( K ) going to be equal to? So pause the video and see if you can figure out ( K ), and I'll give you a hint: you just have to leverage some of your exponent properties.

All right, let's work this out together. The first thing I'd want to do is be a little bit consistent in how I write my exponents. Here I've written it as ( -65 ) power, and here I've written it as a fifth root. But we know that the fifth root of something, we know that the fifth root of ( V )—that's the same thing as saying ( V^{\frac{1}{5}} ).

The reason I want to say that is because then I'm multiplying two different powers of the same base, two different powers of ( V ), and so we can use our exponent properties there. So this is going to be the same thing as ( V^{-65} ) times ( V^{\frac{1}{5}} ), which is going to be equal to ( V^{K} ).

Now, if I'm multiplying ( V ) to some power times ( V ) to some other power, we know what the exponent properties would tell us. I could remind us—I'll do it over here: if I have ( x^{a} \cdot x^{b} ), that's going to be ( x^{a + b} ).

So here I have the same base ( V ). Therefore, this is going to be ( V^{(-65) + \frac{1}{5}} ).

So ( V^{-65 + \frac{1}{5}} ) is going to be equal to ( V^{K} ).

I think you might see where this is all going now. So this is going to be equal to ( V ). Therefore, ( -65 + \frac{1}{5} ) is going to be equal to ( K ).

Calculating this gives us ( -\frac{325}{5} + \frac{1}{5} = -\frac{324}{5} ).

Now, all of this is going to be equal to ( V^{K} ), so ( K ) must be equal to ( -\frac{324}{5} ).

And we’re done! ( K ) is equal to ( -\frac{324}{5} ).