A Nice Algebra Question | How To Solve This Olympiad Problem?

Added: 2026-05-11

[00:00:00]Hello, my friend. Happy to see you here.

[00:00:01]Today, we have a great algebra question.

[00:00:04]We have 2 to the a plus 4 to the a equal to 8 to the power a. Great, quick, and relaxing algebra question. Right now, you can easily pause the video and try to solve this question on your own, and then we will check our answer. So, it will be really interesting. So, right now right now, let's start. First of all, 2, 4, and 8 are great numbers because, for example, 4 we can write as 2 squared, 8 we can write as 2 cubed. So, then we will have the same base. So, let's start with this with this with this moment. So, we have 2 to the a. Yeah, we have 2 to the a, and instead of 4, we write 2 squared. So, 2 squared to the power to the power a equal to instead of 8, we write 2 cubed. So, 2 cubed to the power to the power a. And right now, we need to remember we need to learn a really interesting property, maybe the most popular property when we have a to the power x raised to the power y. So, when we raise this with parentheses, we can write it as a to the power x times y. a to the power x times y. So, we can multiply our our powers. So, I'm going to start it on the left side. So, we have 2 to the a plus 2 to the power 2a. 2 to the power 2a and equal to 2 to the power 3a. 2 to the power 3a.

[00:01:22]Really great. So, what we're going to do next? Every time we have this type of equation, so we have something on the left side, on the right side. My quick recommendation, just write everything on the on the left side. So, that's why we write 2 to the a plus 2 to the power 2a and minus this 2 to the power 3a and equal to equal to 0.

[00:01:42]Okay, so I hope you understand this step.

[00:01:45]What we're going to do next? Right now, let's factor 2 to the a. We can easily factor one of these 2 to the power a.

[00:01:50]So, let's do it. So, we have 2 to the power a and in parentheses from here, we have 1 plus 2 to the power a 2 to the power a and minus 2 to the power 2a because we take away this this one. So, we have 2 to the power 2a equal to 0.

[00:02:08]Right now, what we're going to do next?

[00:02:09]I'm going to I recommend you to write the substitution. So, let 2 to the power a equal to t. So, 2 to the power a equal to equal to t. And we have uh restrictions right here because if 2 to the a equal to t, then t is not equal to 0. t is not equal to 0. And second one, t is greater than 0. t is greater than 0. Okay, so right now, let's use the substitution. As a result, what do we have? t and in parentheses, we have 1 plus t and minus t squared. 1 plus t and minus t squared. And a lot of students might be asking why right here we have t squared because we can write this one as 2 to the power a squared.

[00:02:53]So, maybe you don't understand why I write this as t squared. So, this one we can write So, let me just put the right here this sign. Okay, right here, we can write this as 2 to the power a squared according to the same the same logic.

[00:03:06]Okay? This one is 2 to the power a squared.

[00:03:09]And of course, equal to 0.

[00:03:11]Right now, a product of two parentheses equal to 0 when the first parenthesis or not like parenthesis, we have two expressions. This expression and parenthesis.

[00:03:20]Okay, so a product of two things equal to 0 when the first thing equal to 0.

[00:03:25]So, t equal to 0 or the second parenthesis equal to 0. So, we have 1 plus t and minus t squared equal to equal to 0.

[00:03:37]Okay, right now, t equal to 0, but just look closely right here. t cannot be equal to 0 because then we will have 2 to the a equal to 0. So, exponential function doesn't intersect with with this linear function t equal to 0. So, as a result, we can easily reject. So, t equal to 0, we reject reject this branch, reject this. Maybe right here we have complex roots, but in our case, we solve it for real number roots. Okay, so our last hope is this quadratic equation. First of all, we can change an order real quick. We can write minus t squared on the first position plus t and plus 1. Plus 1 equal to equal to 0.

[00:04:16]And right now, I really like this trick, this quick hint for everyone because we don't like this negative sign in the beginning. So, that's why my quick recommendation, just divide or multiply both sides. No, you you don't need to to write it. Just do it in your in your head.

[00:04:32]Oh, sorry.

[00:04:34]Just do it in your head. Yeah, and right now, let's multiply both sides by minus 1. So, as a result, we have t squared.

[00:04:41]So, we just change all the signs to the opposite one. t squared minus t and minus 1.

[00:04:47]Equal to 0. So, we have a basic quadratic equation.

[00:04:51]So, what we're going to do next? Right now, let's write all these coefficients and let's solve this quadratic equation.

[00:04:56]So, I'm going to do it on the right side. So, a equal to the first coefficient equal to 1. The second coefficient b equal to minus 1. And coefficient c equal to minus 1 as well.

[00:05:07]So, right now, let's solve it for for discriminant. Yeah, let's do it. So, d equal to b squared minus 4 a c equal to b squared. So, minus 1 squared minus 4 times a times 1 and times c times minus 1. So, let's evaluate it. Let's simplify it. So, minus 1 squared equal to 1.

[00:05:29]This negative and this we have plus, so plus 4 times 1 times 1 equal to 4. So, as a result, we have 5. So, our discriminant is positive. This is also really great because with right here at this branch we will have two two real number roots. Two real number number roots. It's also really great.

[00:05:49]So, right now, let's solve it for So, for example, imagine this is our t first. So, let's write it t first equal to 0. Of course, we reject it. Maybe right here we have complex roots. So, that doesn't matter. So, right here at this place, I'm going to write t second and t third. Okay? This is also maybe complex part right here.

[00:06:06]So, t first. But in this case, I'm going to write t second and t third because this is also a root, but a complex one.

[00:06:12]Okay?

[00:06:13]t second and t third equal to So, minus b all known formula plus minus square root of discriminant and all over all over 2a equal to So, minus b minus minus 1 plus minus square root of discriminant square root of 5 over over 2. And as a result, our root we have 1 plus minus square root of 5 >> [snorts] >> over 2. Right now, a few thoughts about this t second and t third because just just learn just remember that t is not equal to 0 and t is greater than 0. Of course, we we also reject reject it one one of these. So, we reject this t equal to 0, but t is also need to be positive.

[00:06:55]And right here, I feel that one of this root will be negative. So, let's let's check it. So, as a result, we have t second. Let's go with the plus sign. 1 plus square root of 5 over 2 and t third equal to 1 minus square root of 5 over 2. And I really [snorts] hope you understand that if you look closely at t third right here. So, 1 minus Of course, 2 is positive, obviously. Yeah, and our denominator is positive. What about this numerator? 1 minus square root of 5.

[00:07:27]Uh it's it's really easy to check it because square root of 4 equal to 2. So, as a result, 1 minus 2 is negative. And even we have square root of 5 is greater than square root of 4, you know? And right here, we have a negative a negative value. So, as a result, our t third is less than 0. So, it means that we also reject reject this t t third. Maybe right here we have complex roots. So, this is up to you if you want to practice your complex numbers, you can easily solve this this branch. But in terms of real numbers, we reject this one. So, our last hope, this is t second.

[00:08:07]Numerator is positive, denominator positive. So, everything is great. So, let's go back.

[00:08:12]Let's go back to our substitution. So, let me just write this thing. So, but uh we had a substitution. So, 2 to the a equal to t. So, but 2 to the a equal to equal to t. So, that's why our next step is to solve the equation 2 to the a equal to t equal to this 1 plus square root of 5 over over 2. So, let's solve it. First of all, maybe we're applying log. Maybe this is the best thing, maybe the only one thing because we need to solve it for a. And the best way to do it is to apply log to both sides. log 2 to the a >> [snorts] >> equal to log 1 plus square root of 5. 1 plus square root of 5 over over 2. Okay, so we have something like that. Right now, this a according to this log property will come down on the in the beginning. So, we have a log 2 a log 2 equal to What What can we do with this with this thing? First of all, I want to leave it like that without any changes. So, 1 plus square root of 5 over over 2. And the final step, this is a constant. This is also a constant. So, if you want to solve it for a, you need to divide both side by log 2.

[00:09:26]Yeah, we need to divide both side by uh by log 2. So, we have our answer log this one 1 plus square root of 5 over 2. And we divide it by by log 2. And a few thoughts about this expression because we can easily simplify this numerator. Yeah, we can easily do this because when we have a log of a fraction, we can write it as a subtraction of of the or like difference of two logs. So, as a result, our a equal to So, log of a fraction, we can write as log this numerator minus log of the numerator. So, log 1 + square root of 5, yeah? So, 1 + square root of 5 - log 2 - log log 2 over over log 2.

[00:10:13]Or we can easily see this log 2 and right here we can easily see this log 2.

[00:10:17]So, as a result we can divide by this log 2. So, this is the common thing. So, we have A equal to So, log 1 + square root of 5 1 + square root of 5 over log 2 - log 2 over log 2 and right here we can easily cancel this log on the right side. So, the answer to this question is A equal to log 1 + square root of 5 over log 2 over log 2 and right here we have 1 so - 1. Or if you interested in approximately answer X approximately equal so, I'm going to write it right here. So, X approximately equal if you plug in it in a calculator 0.69 4. This our approximately approximately answer and this is our our real real number root. So, this my solution to this question. I really hope you understand it. I really hope you learn something new but definitely don't feel bad if you got this wrong. If you need help with any of this classes I have a lot of questions on my YouTube channel a lot of different challenges so, I hope you will enjoy it. So, thank you for your time. Have a great day and see you in the next videos.