two to the x squared minus three power

is equal to one over the cube root of x.

Pause the video and see if you can solve this.

Well you probably realize that this is not so easy to solve.

The way that I would at least attempt to tackle it is,

you would say this is two to the x squared minus three

is equal to x to the,

I could rewrite this, this is one over x to the 1/3,

so this is x to the negative 1/3 power.

Maybe I can simplify it by raising both sides

to the negative three power.

And so then I would get,

if I raise something to an exponent,

then raise that to an exponent,

I could just multiply the exponents.

So it would be two to the negative three x squared

plus nine power.

I just multiplied both of these terms times negative three,

is equal to x to the negative 1/3 to the negative three.

Negative 1/3 times negative three is just 1,

so that's just going to be equal to x.

So it looks a little bit simpler, but still not so easy.

I could try to take log base two of both sides

and I'd get negative three x squared plus nine

is equal to log base two of x.

But once again not havin' an easy time solve this.

And the reason why I gave you this equation

is to appreciate that some equations

are not so easy to solve algebraically.

But we have other tools, we have things like computers,

we can graph things and they can

at least get us really close

to knowing what the solution is.

And the way that we can do that

is we can say hey, well what if I had one function

or one equation, that was y is equal to two x,

two to the x squared minus three I should say.

And you had another that was y is equal

to one over the cube root of x.

And then you could graph each of these

and then you could see where they intersect.

Because where they intersect

that means two to the x squared minus three

is giving you the same y as one over the cube root of x.

Or another way to think about it is they're going

to intersect at an x value,

where these two expressions are equal to each other.

And so what we could do is we could go

to a graphing calculator, or we could go to a site

like Desmos and graph it and at least try

to approximate what the point of intersection is,

and so let's do that.

So I graphed this ahead of time on Desmos.

So you can see here, this is our two sides of our equation

but now we've expressed each of them as a function.

Right here in blue we have two, we have f of x,

or I could even say this is y is equal to f of x which

is equal to two to the x squared minus three.

And then in this yellowish color I have y

is equal to g of x which is equal

to one over the cube root of x,

and we can see where they intersect.

They intersect right over there.

And we're not going to get an exact answer,

but even at this level of zoom

and on a tool like Desmos you can keep zooming in

in order to get to get a more and more precise answer.

In fact you can even scroll over this

and it'll even tell you where they intersect.

But even if we're trying to approximate,

just looking at the graph.

We can see that the x value, right over here

it looks like it is happening at around,

let's see this is 1.5, and each of these is a tenth.

So this is 1.6 and then it looks like

it's about two thirds of the way to the next one.

So this looks like it's about,

I'll say this is approximately 1.66.

And if you were to actually find the exact solution,

you would actually find this awfully close to 1.66.

So the whole point here is, is that,

even when it's algebraically difficult

to solve something you could set up

or restate your problem or reframe your problem

in a way that makes it easier to solve.

You can set this up as, hey let's make two functions,

and then let's graph them and see where they intersect

And the x value where they intersect well

that would be a solution to that equation,

and that's exactly what we did right there.

We're saying that hey, the x value,

the x solution here is roughly 1.66.