Can You Solve This? | Math Olympiad Challenge

Added: 2026-05-10

[00:00:01]Let's find the value of X in this nice exponential equation. Solution here.

[00:00:08]Can I write what we have as this 3 raised to power X times 9 right plus Now here we can write it as 3 squared all raised to power X.

[00:00:20]Plus also here we have 3 cubed all raised to power X equals to 39 from here.

[00:00:29]That is when we apply the law of indices A raised to power M all raised to power N same thing as A raised to power MN it is same thing as A raised to power N raised to power M.

[00:00:42]That is A. We have 3 raised to power X plus this will go 3 raised to power X all squared.

[00:00:51]Plus 3 raised to power X all cubed equals to 39 from here.

[00:00:58]Then here we have 3 raised to power X is common. Can let a letter Y be equals to 3 raised to power X which implies this equation becomes Y plus Y squared plus Y cubed equals to 39.

[00:01:20]Then also we can express 39 here as 3 plus 9 plus 27.

[00:01:28]Now what we have becomes Y plus Y squared plus Y cubed equals to 3 plus 9 plus 27.

[00:01:42]That is can rewrite this as Y plus Y squared plus Y cubed equals to 3 plus 9 that's 3 squared plus 27, that's 3 cubed.

[00:01:59]And we bring everything to one side.

[00:02:02]We arrange. Here is our Y minus 3 plus Y squared minus 3 squared plus Y cubed minus 3 cubed equals to 0 here.

[00:02:19]We sub same thing as Y minus 3 into brackets plus Y squared minus 3 squared into brackets and plus Y cubed minus 3 cubed into brackets equals to 0 here.

[00:02:37]And this first bracket the second bracket here follows what we have, A squared minus B squared which is same thing as A minus B into brackets.

[00:02:49]So, open brackets A plus B.

[00:02:52]This bracket follows what we have, A cubed minus B cubed which is same thing as A minus B into brackets and open brackets A squared plus AB plus B squared.

[00:03:07]That is we can rewrite this equation and we have Y minus 3 into brackets plus here becomes Y minus 3 into brackets open brackets Y plus 3 close brackets also plus here we have Y minus 3 into brackets open brackets Y squared plus 3 Y plus 3 squared close brackets equals to 0 here.

[00:03:39]Then, here we have Y minus 3 common. We factor it out here, Y minus 3 into brackets open brackets. Here we are left with 1 and plus Y plus then plus here we have y squared plus three y plus nine close brackets equals to zero here.

[00:04:05]Next step, we can write this as y minus three into brackets and open brackets.

[00:04:12]Here we have y squared and y plus three y plus plus four y one plus nine 10 plus three plus plus 13 close brackets equals to zero here.

[00:04:25]Here we have two possible cases.

[00:04:28]First one we have y minus three equals to zero or we have y squared plus four y plus 13 equals to zero.

[00:04:41]Solving here, we have y equals to three.

[00:04:45]We can recall that we have represent y as three raised to power x which implies we have three raised to power x equals to three which can be written as three raised to power x equals to three raised to power one.

[00:05:05]Then we equate the power here from the law of indices which is a raised to power m equals to a raised to power n implies m equals to n.

[00:05:16]And here we have x equals to one which is a real solution here.

[00:05:23]Then also on this side we check if we have a real solution here.

[00:05:32]When our discriminant b squared minus four ac is greater than zero, we have a real solution.

[00:05:40]Otherwise, no real solution. So here a equals to one b is equals to four And C equals to 13.

[00:05:51]That is the discriminant becomes 4 squared minus 4 times 1 times 13.

[00:06:00]We have discriminant giving us 16 minus 4 times 1 times 13 equals 52.

[00:06:09]That is the discriminant here we have 52 minus 16.

[00:06:15]Thus 2 3 plus minus 32 which is less than 0.

[00:06:28]Which means we have no real solution here.

[00:06:35]Okay, so the real solution we have here is x equals to 1.

[00:06:39]Therefore, the value of x in this given problem same thing as x equals to 1.

[00:06:51]And when we substitute into what we have the value of x is becomes 3 raised to power 1 times plus 9 raised to power 1 plus 27 raised to power 1 is equals to 39.

[00:07:08]This is 3 plus 9 plus 27 is equals to 39. Of course, 3 plus 9 plus 27 equals 39 which is equals to 39 here.

[00:07:21]We have left hand side equals to the right hand side.

[00:07:26]And therefore, this value of x satisfy this given problem. Thank you for watching. Don't forget to subscribe for more videos.

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[00:07:38]See you in the next class and bye for now.