USA Mathematics Trick That Got Many Students Excited

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[00:00:06]All right.

[00:00:09]Okay.

[00:00:11]I'll try to go live here again.

[00:00:17]Okay.

[00:00:19]So, we have to do our necessary setting first before we start off here.

[00:00:26]All right.

[00:00:27]Welcome to today's class. Here we have this exponential challenge here before us here. The question is 2 to the power of x + 1 plus our 2 to the power of x - 1 equal to 20. What is the value of x here? So, let's take our solution here.

[00:00:48]All right.

[00:00:50]Now, the trick we want to apply here in solving exponential equation of this kind is this.

[00:00:56]Here we are having 2 to the power of x + 1. So, these exponent here and these exponent here are different. So, what we want to achieve here, which is the trick to solving exponential equation of this kind is that we'll either bring out this x + 1 from here or we bring out x - 1 from this power here. But, being the first one, so let's bring out x + 1 from this second power here. So, our expression will have 2 to the power of x + 1 plus our 2 to the power of x. Now, our x - 1 will be written as our x + 1 here, then our 2.

[00:01:49]Because we have + 1 - 2 will give us -1.

[00:01:54]So, this has not changed anything. So, this is equal to our 20.

[00:02:00]Remember the trick. The trick here is to achieve what we have here at this other side also.

[00:02:07]So, I can put this in bracket.

[00:02:09]Put this in bracket.

[00:02:11]Because if I open up this, we're still going to have our expression on this other side. So, this is going to give us our two to the power of x + 1, then plus our two to the power of our x + 1.

[00:02:27]Here, I will apply a law in indices which says that when you have your uh let's take this. When you have your uh a to the power of m minus n, this is equal to your a to the power of m times a to the power of minus n.

[00:02:51]Right? Good. So, we can now write this as times our two to the power of our minus two. Everything equal to 20.

[00:03:02]Wow.

[00:03:03]And so, from here, what we do, you discover that we have two to the power of x + 1 here. We equally have two to the power of x + 1 here. So, we can factor two to the power of x + 1 from the system. So, this is going to give us our two to the power of x + 1 bracket.

[00:03:23]We use this to divide this. We are left with one here. Plus, we use this to divide this. We are now left with our two to the power of minus two.

[00:03:32]All in bracket equal to our 20.

[00:03:35]Easy.

[00:03:37]Good. We can rewrite this applying the law in indices which says that your a to the power of minus m is same thing as one all over a to the power of m.

[00:03:51]Right? So, with this, we can rewrite this. So, is going to give us 2 to the power of X plus our one bracket our one plus our one all over 2 to the power of 2 close bracket equal to our 20.

[00:04:06]Now, with this we have here, 2 to the power of 2 is 4. So, we're going to end up having 2 to the power of 2 plus sorry, X here plus our one bracket our one plus one all over four close bracket equal to our 20.

[00:04:26]Okay, so let's continue on this side and see what this gives us here.

[00:04:30]Let's erase this first. Now, look at this here.

[00:04:33]This is all over one. So, if we find the LCM and we manipulate this, we're going to end up having 2 to the power of X plus our one, right? Bracket our Yeah, the LCM is four. So, if we use four to divide one to divide four, it's going to give us our one. One times four rather, we use one to divide four will give us our four. Therefore, times our one will give us four. So, we're going to have four plus Yeah, we use four to divide four will give us one. One times four will give us our one here all over our four close bracket everything equal to zero.

[00:05:13]So, we can add these to give us 2 to the power of X plus one bracket our five all over four that equal to 20. Easy.

[00:05:24]Now, we want to eliminate this 20 here.

[00:05:27]So, what I want to do here is to I want to multiply both side of the equation by four all over five.

[00:05:33]And so, this is going to give us 2 to the power of X plus one times Yeah, we have our five all over four already there times four all over five equal to 20 all over one times our four all over five.

[00:05:53]And so this will go with this, this will go with this. On this side we are just left with our plus one equal to the five here one, five here we have four. So we going to have four times four. And four could be written as our two to the power of two there times two to the power of two.

[00:06:18]Simple.

[00:06:20]Because we are having a multiplication sign here, we can take one of them and then add the exponent. Applying another law in indices which says that your a to the power of your m times your a to the power of your n, this is equal to your a to the power of your m n.

[00:06:42]And so this going to give us our two to the power of two plus two. So we going to end up having two to the power of x plus one equal to our two to the power of four.

[00:06:54]The bases are the same and so we equate the exponents. So we have x plus one equal to our four. Subtract one from both sides. We going to have x plus one minus one equal to four minus one. So this this leaves. So now have therefore our x is equal to our three.

[00:07:14]So three is the value of x that satisfy this exponential equation. Applying this trick, I believe it saves a lot of time.

[00:07:25]But I want to take another challenge, another example that is in a different format but same exponential equation.

[00:07:32]And how do we solve that also? All right, so let's erase this.

[00:07:38]Okay.

[00:07:40]Uh All right.

[00:07:54]Now, so our question is in this format this time around.

[00:08:00]So, we have our forgetting our 2 to the power of 2 x times our 9 to the power of x equal to 1 all over 6. These are simple exponential equations, though.

[00:08:15]All right, so when you have a question of this, at the end of the video we're going to do a verification to the root to this very one. So, we are asked to look for x. So, what we do here is just to rewrite this in the simplest base, which is base three. And this cannot be simplified beyond this. And so, this going to give us our Let's take our solution first.

[00:08:38]Our solution, so we have our 2 to the power of 2 x times 9 could be written as 3 to the power of 2 all in bracket raised to the power of x equal to 1 all over 6.

[00:08:51]We apply a law in indices, which says that when you have your a all raised to the power of m times n, this is equals to your a to the power of your m bracket raised to n is equivalent to your a to the power of your n all in bracket raised to m.

[00:09:11]So, if we apply this law, then we can use this to multiply this.

[00:09:16]So, this going to give us 2 to the power of 2 x times our 3 to the power of 2 x equal to 1 all over 6.

[00:09:25]Wow. Now, look at what we have in here now.

[00:09:28]The powers are the same, but the base numerals different. Can we multiply? Yes.

[00:09:35]According to the law in indices, which says when you have your A to the power of your M times your B to the power of your M, so this is equal to your A B all raised to the power of M.

[00:09:51]Right? So, if it is true, then what happened?

[00:09:55]We just have to multiply this. This is going to give us our two times our three all raised to the power of two X this is equal to one all over six.

[00:10:08]So, two times three will give us six.

[00:10:11]So, this going to give us our six raised to the power of two X equal to our one all over six.

[00:10:18]Good. Now, we are having six here, but we have one all over six on this other side. So, what we do here?

[00:10:25]We just have to rewrite this. Applying the law, which says that your one all over A to the power of M is equal to your A raised to the power of minus M.

[00:10:40]So, we want to take this is in this format, we want to take it to this format. So, rewriting this, this going to give us our six to the power of two X is equal to six raised to the power of minus one.

[00:10:53]Check out the base numbers. They are the same, so we equate the exponents, right?

[00:10:58]Okay. So, this going to give us our two X will not be equal to minus one.

[00:11:06]Since we are looking for X, we just have to divide by two. Divide this by two.

[00:11:10]This this cancel out. So, we have X is equal to minus one all over two.

[00:11:16]So, the next question here now is this.

[00:11:19]Does this root satisfy the original equation, or is it extraneous? So, let's do a simple substitution and check if it satisfy the original equation.

[00:11:29]So, So, take our check. The original equation is 2 to the power of our 2x * 9 to the power of our x equal to 1 all over 6.

[00:11:42]So, what does it implies? It means that whenever we see x, we've got to put in -1 all over 2. And so, the first one you're going to have 2 to the power of 2 * -1 all over 2.

[00:11:54]Then times our 9 raised to the power of -1 all over 2.

[00:12:00]This is equal to 1 all over 6.

[00:12:03]Remember the law in indices, which is this law here. So, if I apply this law, what happened? This will go with this.

[00:12:12]We're now left with 2 to the power of -1.

[00:12:16]times Now, we still have this. Let's just keep this. So, we have our 9 to the power or let's go ahead and express Okay, let's just go first. So, we have this.

[00:12:29]Then this is equal to 1 all over 6.

[00:12:32]So, let's continue on this side.

[00:12:35]Let's express this in its simplest base.

[00:12:38]Okay? So, this is going to give us our 2 to the power of -1 * 3 to the power of our 2 then times -1 all over 2.

[00:12:52]Okay? Then this is equal to our 1 all over 6.

[00:12:57]Remember applying the law of indices again, which is this law here, that will leave So, now I have 2 to the power of -1 times our 3 to the power of -1 equal to 1 all over 6.

[00:13:10]We can rewrite this expression applying this law.

[00:13:14]Yeah. Okay, so if I apply that law, that is going to give us our 1 all over 2 to the power of 1 times our 1 all over 3 to the power of 1 equal to our 1 all over 6.

[00:13:28]So, 2 to the power of 1 is 2 and 3 to the power of 1 is 3. And so, we can multiply out. And so, this is going to give us our one all over 2 * 3 will give us 6. And on the right-hand side, we have one all over six.

[00:13:42]So, this shows that the answer or the root we solved for here, x equal to minus one all over two satisfy the original equation.

[00:13:52]So, these are some of the simple tricks um methods you apply in solving simple or exponential equations of this kind.

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