Are YOU smart enough to get into Cambridge?

Hello my lovelies. It's Susanna and today I want to show you how to solve this problem. It was a question in an admission test for the University of Cambridge.

In this test you're not allowed to use a calculator and you have to solve 20 questions within 75 minutes. So on average around four minutes per question. So not a whole lot of time.

So let's see if we can solve this within four minutes. Find the sum of the real solutions of the equation that looks like this. We get these six answers here.

So we have to solve this equation and then find the sum of all the real solutions that we found. So if this equation has more than just one solution, then we just add all these solutions and pick the correct answer here. Okay, let's start and try to solve this equation.

It's an exponential equation because we have x in the exponent here and also here. So we have 3 to the^ of x and the<unk> of 3 to the power of x + 4. So we have to use the logarithm at some point to be able to solve for x.

But it's not a good choice to apply the logarithm right now because we have a minus here and a plus here and the logarithm wouldn't help us to get any um solution. So we first try to look at the equation try to simplify it a little bit. For example, here we have this sum in the exponent.

We could write this in a different way. So every time you have a sum in the exponent. So something like x to the^ of a + b the sum here.

Then there is a rule that you can write it as two separate powers. So you take your base the x and raise it to the power of the first part of the sum. So to the power of a and then you multiply it by the base again and then you raise it to the power of the second part of the sum to b.

Let's apply this rule to this expression here. So we have 3 to the^ of x minus. Now we take the base in our case it's the square root of three.

we write it in parentheses and then we raise it to the first part of the sum. So here it is the x we raise it to the power of x and then we multiply it by we take the base again. So this wonderful expression of the square root of three and then we raise it to the second part of the sum.

So we raise it to the power of four. Then we have the rest of the equation and take a look at what we can do next. Maybe now we can calculate this expression here because we just have the square root of 3 to the^ 4.

We can calculate this and write it in a nicer way than this. So we have all the things here. We have the square<unk> of 3 and to the power of x.

We can leave it like this. But now what is this expression? We have the square<unk> of 3 raised to the power of 4.

There are several ways how to calculate this now. But one way is to write the square root as a power. So we have all of this raised to the power of four and instead of the square root of three I only take the three and the square root can always be written as raised to the power of 1 over two.

So instead of a square root you can always just raise this thing to the power of 1 /2 because now we can just apply the rule how to simplify this. We just keep the base the three and multiply the exponents. So we have 1 / 2 ultiplied by 4.

And if we calculate this we have 4 over 2 which equ= 2. So we have 3 ^ 2 which equals 9. So this is the result for this expression here.

And then we just write down the rest of the equation. And everything looks better already. I usually don't like it if the numbers are not in front of everything with my x.

So I would switch these two expressions here. So, I take the first thing, then I put my nine in front of this expression with my x and then I write down the rest of the equation. So, this is what I usually like.

Okay, this is how our equation looks like. Now, if we take a look at the exponents, we have x as an exponent here and x as an exponent here as well. So they are the same which is great already.

But if we take a look at the bases, we have base three here but the square root of three as a base here. They are not yet the same. Similar but not the same.

But we just saw that we can write the square root of three as a power. Right? We know that we can write this as 3 to the power of 1 over two.

Instead of the square root, I just write it like this. And then I have my base three that I was looking for. So let's write this expression here as this.

We have 3 to the^ of x. We subtract nine times. Instead of the square root, we write this power here and raise it to the power of x.

And then we write down the rest of the equation. Okay, we have base three here. We have base three here, which is great.

Um, but let's simplify this expression here the same way we did. We can just multiply the exponents. So we have 3 to the^ of x.

So we subtract nine times. Now we take the base the three and we multiply the exponents. We have 1 over 2 * x then as our new exponent and the rest of the equation.

So what is so special about this equation now about the structure of this equation? We have three parts here. In the first part we have base three and raise it to the power of x.

In the second part we also have base three but raise it to half of x and then we have a part where there is no x at all and on the other side we have zero. Every time you have this structure where the bases repeat and there is a connection between the exponents, right? So we have x here and half of x here.

So this exponent is twice as large as this exponent. Every time you have this structure, you can use substitution to solve this equation for x. So also if your uh equation would look differently, for example, if you had x ^ 6 - 9 * x ^ 3 + 20 = 0.

Same thing here. Let's compare it. We have the base x here, base x here.

So same bases another part where there is no x in there at all and the exponents there is a connection between the two. This number is twice as large as this number. So also for this equation for example that looks completely different but has the same structure you could also use substitution to solve it.

So it's a common way to solve equations. If you see the structure, how can we use substitution now to solve this equation? Substitution means you take a look at your equation and look for the things that make your equation complicated.

So here obviously 3 to the^ of x makes my equation complicated and this expression here that is the similar expression makes it also complicated. If they would look differently the equation would be much simpler. So we're going to substitute these complicated things now by something that is way simpler.

We take this thing here and from now on we don't write it down like this. We call this whole orange expression from now on just U. You can pick any letter that you like.

Sometimes it's a Z, sometimes it's whatever. You can choose whatever letter you like. We take U here.

So instead of this expression, we're going to write just U. We substitute this thing by U. But we also have to substitute this thing here but not by another letter.

So we don't say this is v. But we want to have it in terms of u as well. And that's why it was so important that you have a connection between these two expressions that the bases are the same and that the exponents have a connection that this exponent is twice as large as this one.

Because if that's the case, then this thing here is just u squared always. If this is your u, then this thing here is your u squared because this exponent is twice as large as this one. And if you don't believe me, we can just prove it.

So just take your u expression. So let's do it here. That we have 3 ^ 1 / 2x.

And if we square it now, we want to get this as a result so that we know that this works out here. So let's take this let's square it and see what we get. We can take the base the three and just multiply the exponents.

So we have 1 over 2x and multiply it by two. And then we see that the two cancels out. Here we have three as a base and our exponent is just the x.

So we've just seen okay that's the result here if we square our u. So the things work the connection is great between our expressions and the substitution works. So my substitution is instead of this expression I'm going to write a u.

Write it down what you substitute. So we have this here uh and we call this u from now on and we're going to need this connection later on. This is our substitution.

So in terms of u how does my equation look like? Then instead of this thing here, we said we write u squared from now on. We subtract nine times.

Instead of this beautiful thing, we write a u from now on. And then the rest of the equation. So this equation is way simpler than this one.

And this is the great thing about substitution. It makes things easier first. So the plan now is to solve this equation for u.

And if we have our u solution, whatever it's going to be, we were not really interested in solving this equation here, right? We we were looking for solutions for this original equation. But if we have our equa um solutions for u, we can go back to our substitution here and use this equation to find our solutions for x in the end.

Then so that's how substitution works. You make things easier first. you solve easier equations and go back to your substitution and find your solutions that you were looking for in the end.

Okay, a lot of things to do. So, take this equation and let's solve this for you. It's a quadratic equation.

So, we could use the quadratic formula to solve this. Or because I know many of you prefer factoring if it's possible. So, if these numbers here are nice enough, we could also just factor the left side of this equation.

Let's try this to write the left side as a product. And on the other side, we have a zero. Because we have u squared here, we take the u and write it in the parentheses here and also here.

And now we're looking for numbers to fill in here and here. And if we multiply these two numbers, so let's call them a and b. If I multiply a by b, then I have to get this number here as a result.

So the 20. And if I add these two numbers that I still have to find. So if I add a and b then I have to get this number as a result.

So the -9. So I'm looking for two numbers now that if I multiply them I get 20 as a result. This would be something like 4 * 5 for example.

And then I always check if the second condition is fulfilled as well. So if I add these two numbers. So if I do 4 + 5, do I get -9 as a result?

No. But I get nine as a result. So almost.

But I also know that my numbers need to be negative so that I get a negative result here. So it can't be four and five. But if I multiply -4 by -5, I also get 20 as a result.

But if I add these two negative numbers now, I get -9 as a result. So these are my numbers I was looking for. I have the -4 that goes in here and the -5 that goes in here.

And now I can solve this equation pretty fast because I have a product equals zero. This means that this is only possible if either my first thing here in my parentheses equals 0. So either u - 4 = 0 or as a second equation my second parentheses equals 0.

So u - 5 = 0. To solve this for u, I just add four on both sides of the equation so that I have my u here. This cancels out and on the other side I have my four.

A first solution by the way. And a second solution for my u. I add the five here on both sides of the equation so that I have my u on one side because this cancels out and on the other side I have the five.

Another solution. I have two solutions for my U. I call this U1 and this U2.

I found two solutions for you. But I was not interested in u solutions, right? We were interested in solutions for our x equation here.

But we said we will use these u solutions. now and use our substitution to find our x solutions. So I insert this solution now for my u and solve this equation for x and then I do the same with the second solution.

But first the four. So I put the four in for my u and here I have 3 ^ 1 / 2x. And now we're going to solve this equation for x.

Let's do this on this side here. We want to solve for x. We have x in the exponent of our equation.

And on the very beginning of this video, a few minutes ago, I told you that we have to use the logarithm to be able to solve exponential equations. And now is the perfect time to use the logarithm because this equation is simple enough to use it on both sides. So I want to apply the logarithm on both sides of this equation.

And the easiest way to use the logarithm is to use it to this base. So log base 3 is what I'm going to apply on both sides of the equation now. So I have log base 3 of 4 on the left side.

The same on the right side. I have log base three of this whole thing here. So I have to write everything inside my logarithm.

This looks a little bit more complicated than this. I admit that. But it's really good as it is right now.

We just leave the left side as it is because this is just a number. There is no x in here. But here we can use a rule for the logarithm.

If you have an exponent in your logarithm, you are always allowed to write this exponent in front of your logarithm. So you take your 1 over 2x and write it in front of the logarithm. And then you just have your logarithm base 3 off.

And then you only have the three left in here because everything else is in front of the logarithm already. So you only have the three in your logarithm. And now this expression here is easy to calculate because every time the base and the number in your logarithm are the same.

Then this logarithm just equals one. So we have this times one here. So instead of this log I just write a one here.

Times one. I don't have to write it at all. So this looks like this.

And to solve this equation for x now I just have to get rid of the 1 / two here. So I can just multiply both sides of the equation by two. So here times two and here times two.

On the left side I have this number here. Don't worry that we still have this log here. We don't have a calculator anyway.

So we can't calculate this. We just take it as it is. And on the other side, these two here cancel out and we only have x left.

So we solved the equation for x. This is our first solution. So I call it x one.

Our first solution here. I wrote it down here again for our original equation. So this u solution gave us an x solution.

And this is what we do with the second u solution as well. So we take the five this time, insert it here for the u and solve this equation for x in the same way. So we are going to be a little bit faster.

Now we know how to solve these exponential equations with base 3. We apply the log base 3 on both sides of the equation. So that we have log base 3 of five on the left side and the same log base 3 of the right side.

So of this whole expression here. The left side is just a number. We leave it as it is.

But on the right side we apply the rule again that we take the exponent and write it in front of our logarithm. So we have 1 / 2x again and multiply this by this log expression. We only have the three left in here again.

And we now know that if the base and this number are the same, this whole thing just equals one. So same thing here that we don't have to write this down here at all. We solve for x by multiplying both sides of the equation by two.

So that we have this number now with log. We don't freak out. We just leave it as it is.

We will see what uh how we can work with it later on. But here the two cancels out and we have our x solution a second solution. So I call it x2.

I wrote down our two solutions here again. And if we take a look at the answers, our solutions are not in there. Why?

Because they were looking for the sum, remember? So, we have to add these two real solutions that we found. So, we take these two and add them and then let's just pray that we will find one of the answers then.

So, let's add these two. Okay, we want to add um two logarithms. It's easier if we only have log expressions, but right now we have a two here in front of my log and a two in front of my log.

So, because I have a sum here, I could factor out the two first. So, let's do that. If we factor out the green two, what is left in my parentheses?

I have the plus here. And here, if I factor out the two, only my log is left, which is good. We wanted to isolate the logs and the same thing here.

I take this out. So only the log is left here. So now I have a sum of two logarithms and the bases of these two logarithms are the same which is great because then we have a rule how to add logarithms.

We can write this as one logarithm. We just take log base 3 and inside our logarithm we now have to multiply these two numbers. So we take the four and multiply it by the five.

And 4 * 5 just equals 20. So we have the 20 here in our logarithm. We don't need these parentheses here then.

So our expression looks like this. Let's pray that it is somewhere in the answers. It's answer E.

And we finally almost in four minutes solved this problem. I'm curious how you solve this problem. So please let me know in the comments.

I wish you a wonderful day and I hope to see you in one of my next videos. Take care.