Only 1% Got This Viral Math Olympiad Challenge! | Can You Solve It?

[00:00:01]Hello, you're welcome. I just solved this nice exponential equation to find the value of x here.

[00:00:08]Solution from here.

[00:00:12]From what we have, which is 2 * root x all raised to power 6 equals to 64.

[00:00:24]First step here, this formula that we have a * b all raised to power n Here, we can write this as a raised to power n * b raised to power n.

[00:00:36]That is, what we have here is the same thing as 2 raised to power 6 * root x raised to power 6 equals to 64 on this side.

[00:00:50]Then, next step here, we can write root x.

[00:00:56]When we have root x, it's the same thing as x raised to power 1/2.

[00:01:02]This implies we have 2 raised to power 6 * x raised to power 1/2 which is raised to power 6 equals to 64 here.

[00:01:15]Then, this power multiplies. When we have a raised to power n raised to power n, I write this as a raised to power m n.

[00:01:25]Then, what we have becomes 2 raised to power 6 * x raised to power 1/2 * 6 equals to 64 here.

[00:01:39]Then, yeah, 2 goes 1, and 2 goes here 3.

[00:01:44]We have 2 raised to power 6 * x raised to power 3 equals to 64 on this side.

[00:01:55]Then, we divide both sides by 2 raised to power 6. That is, divide this by 2 raised to power 6. Also, divide this by 2 raised to power 6.

[00:02:04]Then, here, 2 raised to power 6 cancel each other. We have x raised to power 3 equals to 64 over 2 raised to power 6. 2 multiply 6 six times.

[00:02:16]That is also 64.

[00:02:19]That is, here we have x raised to power 3 equals to This cancel each other. We have 1 left.

[00:02:26]That is, x raised to power 3 equals to 1.

[00:02:30]Next step here, we can rewrite 1 also as 1 raised to power 3.

[00:02:36]And this equation becomes x raised to power 3 equals to 1 raised to power 3.

[00:02:43]Then, we take 1 raised to power 3 to left-hand side. This becomes x raised to power 3 minus 1 raised to power 3 equals to 0 here.

[00:02:55]And this follows from we have a raised to power 3 minus b raised to power 3 is same thing as a + b a minus b other into bracket into bracket a squared plus ab plus b squared.

[00:03:16]That is, here, what we have becomes x minus 1 into brackets into brackets x squared plus x times 1 plus 1 squared close bracket equals to 0 here.

[00:03:37]That is, this [clears throat] is same thing as x minus 1 into brackets into bracket x squared plus x times 1, that's x and plus 1 squared, that's 1 close bracket equals to zero here.

[00:03:54]Then we have two possible cases here.

[00:03:57]First one, x minus one equals to zero or we have x squared plus x plus one equals to zero.

[00:04:10]Then solving on this side, we have x equals to one.

[00:04:15]Which is a real solution here.

[00:04:19]Then also here we have a quadratic equation where a [snorts] equals to one b equals to one and c equals to [snorts] one.

[00:04:31]Applying the quadratic formula which is x equals to minus [snorts] b plus or minus square root of b squared minus four ac all over two a.

[00:04:46]Then what we have becomes x equals to minus one plus or minus square root of one squared minus four times one times one over two times one.

[00:05:06]Then this becomes x equals to minus one plus or minus square root of one squared is one minus four times one times one that's four over two.

[00:05:20]Then here we have x equals to minus one plus or minus square root of one minus four that's minus three over two.

[00:05:32]Then we can write this as x equals to minus one plus or minus square root of minus three that's three times minus one over two.

[00:05:46]Then this follows when we have root A times B. I'll write this as root A times root B.

[00:05:54]Which implies we have X equals to minus one plus or minus root three times root minus one over two.

[00:06:09]Then here we have X equals to minus one plus or minus root three and root minus one is I, which is a complex number over two.

[00:06:20]But this here we have two complex solutions here.

[00:06:25]And therefore altogether in this problem we have three solutions here. One real solution here and two complex solutions here. When we write it out we have X1 equals to one.

[00:06:38]X2 equals to minus one plus root three I over two.

[00:06:48]And we have also X3 equals to minus one minus root three I over two.

[00:06:59]That is we have three solutions together here.

[00:07:04]Then we can check if this satisfy this given problem by substituting the value of X here into this problem. This is two times root X all raised to power six equals to 64.

[00:07:23]When X equals to one this becomes two times root one all raised to power six is equals to 64 here.

[00:07:35]Then here root one is one, so we have two times one all raised to power six is it equals to 64 on this side.

[00:07:46]And 2 times 1 that's 2 which is raised to power 6 is equals to 64 on this side.

[00:07:52]And 2 raised to power 6 here equals 64 which is equals to 64 from here.

[00:08:00]As this we have left hand side equals to the right hand side and therefore x equals to 1 satisfy this given problem.

[00:08:15]We can also check for these two complex solutions. You can try that out and put your answer in the comments. Thank you for watching. Don't forget this step.

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[00:08:25]See you next class and bye for now.