Germany | A Nice Olympiad Algebra Problem

Hinzugefügt: 2026-06-02

[00:00:00]that the 36th of X minus 9 to the power of X, this is equal to 18 to the power of X.

[00:00:07]So, what is the value of X? Now, let's provide a solution from here.

[00:00:15]We can express 36. This is the same thing as 9 multiplying by 4.

[00:00:21]This is raised to the power of X minus 9 to the power of X.

[00:00:28]This is equal to 18 to the power of X.

[00:00:30]We can express 18 as 9 times 2 raised to the power of X.

[00:00:38]Now, we have that 9 * 4 to the power of X.

[00:00:41]This is in the form of A * B raised to the power of N which we can express as A to the power of N times B to the power of N.

[00:00:54]Applying this exponent property, we have 9 to the power of X multiplying by 4 to the power of X subtract 9 to the power of X.

[00:01:05]This is equal to 9 to the power of X multiplying by 2 raised to the power of X.

[00:01:15]Now, let's divide everything here by 9 to the power of X by 9 to the power of X.

[00:01:24]So, let's simplify here.

[00:01:26]So, we have 4 to the power of X minus 9 to the power of X simplifies.

[00:01:32]So, this is minus 1.

[00:01:34]This is equal to 2 to the power of X.

[00:01:41]Now, the next step is that we can express 4 as 2 raised to the power of 2 then raised to the power of X subtract 1. This is equal to raised to the power of X.

[00:01:56]2 to the power of 2 to the power of X, this is in the form of A to the power of N raised to the power of M which we can express as A to the power of M raised to the power of N.

[00:02:10]Applying this exponent property, here we have 2 raised to the power of X raised to the power of 2 - 1 this is equal to 2 raised to the power of X.

[00:02:26]Now, we have 2 to the power of X is common here.

[00:02:29]We can let 2 to the power of X be equal to U.

[00:02:37]Now, let's substitute 2 to the power of X with U. So, we have U to the power of 2 1, this is equal to U.

[00:02:46]Let's take U on the left-hand side. We have U to the power of 2 U - 1, this is equal to to 0.

[00:02:56]We form a quadratic equation here. We have the value of A is equal to 1 B is equal to -1 and C is equal to -1.

[00:03:08]So, we can solve for U by applying the quadratic formula which is -B + or - the square root of B squared - 4 AC divided by 2 multiplied by by A.

[00:03:23]So, let's substitute the values of A, B, and C in this formula here.

[00:03:29]So, this is equal to - * -1 + or - we have the square root of -1 to the power of 2 - 4 * A which is 1 * C which is -1 divided by 2 * 1.

[00:03:49]So, here we have - * -1, this is 1.

[00:03:54]Plus or minus, we have the square root of -1 squared, this is 1.

[00:03:59]-4 * -1, this is +4 divided by by 2.

[00:04:08]So here we have 1 plus or minus the square root of 1 + 4, this is 5 divided by by 2.

[00:04:19]So we have two values of U. We have U1 is equal to 1 plus square root of 5 over 2.

[00:04:29]We also have U2 which is 1 subtract square root of 5 over over 2.

[00:04:41]Now, if you recall we are saying that we let 2 to the power of X be equal to U.

[00:04:52]So in other words, if we have 2 to the power of X, this is equal to 1 plus square root of 5 >> [clears throat] >> divided by 2.

[00:05:02]We have 2 to the power of X, this is equal to 1 minus square root of 5 divided by by 2.

[00:05:13]2 to the power of X, this is greater than 0.

[00:05:17]We have 1 plus square root of 5 over 2, this is also greater than 0. So this means we have positive result here.

[00:05:26]If we check on case two, let's call this case two and this case case one.

[00:05:34]Case two here, 2 to the power of X is greater than 0.

[00:05:38]On the right hand side here, 1 minus square root of 5 over 2, this is less than 0. Because we have the square root of five, this is the same thing as 2.

[00:05:51]236 2361.

[00:05:58]So, this is to mean that 1 minus 2.2361 This is negative. We'll have a negative value here.

[00:06:08]And this means we'll have undefined result. We'll have undefined So, case two here This is rejected. Case two is rejected.

[00:06:25]So, what we have is two to the power of X. This is equal to 1 plus square root of five over over two.

[00:06:36]Let's solve for the value of X here.

[00:06:37]Let's introduce logarithm on both sides.

[00:06:40]Log two to the power of X This is equal to log 1 plus square root of five over over two.

[00:06:52]Log two to the power of X, this is in the form of log A to the power of B which we can express as B log log A.

[00:07:03]Applying this power of property log two to the power of X becomes X log two.

[00:07:10]This is equal to log 1 plus square root of five divided by two.

[00:07:20]Now, let's divide on both sides by log two.

[00:07:24]Here we have log log two.

[00:07:30]So, let's simplify log two and log two here.

[00:07:33]X is equal to log This is 1 plus square root of five.

[00:07:41]We have This is to base to base 2.

[00:07:48]to base 2 minus log 2 over log 2.

[00:07:55]So, if we simplify here, we have 1.

[00:07:58]And this means that the value of X is equal to log 1 + square root of 5 to base 2 subtract 1.

[00:08:10]So, this is the value of X.

[00:08:15]Now, the next step is to verify. Let's verify that this value of X here satisfies the equation. Now, let's verify that this value of X satisfies the equation.

[00:08:30]We have 4 to the power of X minus 1.

[00:08:33]This is supposed to give us a value of 2 to the power of X.

[00:08:39]We can express 4 as 2 to the power of 2 multiplying by X. X is log 1 + square root of 5 to base 2 subtract subtract 1.

[00:08:57]And this is subtracting by 1.

[00:09:00]This is supposed to give us a value of 2 to the power of log 1 + square root of 5 to base 2 subtract 1.

[00:09:12]So, here we have 2 raised to the power of 2 log 1 + square root of 5 to base 2 subtract 2 * 1, which is 2.

[00:09:24]Subtract 1. This should be equal to 2 to the power of log 1 + square root of 5 to base 2 subtract 1.

[00:09:38]So, this is a power here. So, we can express this as two to the power of log 1 + square root of 5 raised to the power of two to base two minus one. This should be equal to two to the power of log 1 + square root of 5 to base two subtract one.

[00:10:03]Now, here we have negative two. We have This is log 1 + square root of 5 to the power of two to base two minus two. So, we have minus two here.

[00:10:14]So, this is in the form of a to the power of n minus m which we can express as a to the power of n over a to the power of m.

[00:10:26]So, applying this property, this means we have two to the power of log 1 + square root of 5 to the power of two to base two divided by two to the power of one. This This is minus by one.

[00:10:44]This should be equal to two to the power of log 1 + square root of 5 to base two.

[00:10:53]Sub Now, this is divided by two to the power of one.

[00:11:00]Now, two to the power of log 1 + square root of 5 to the power of two to base two.

[00:11:05]This is in the form of a to the power of log b to base a and this should give us a value of b.

[00:11:15]So, applying this logarithm property two to the power of 1 + square root of 5 squared to base two this becomes 1 + square root of 5 raised to the power of two.

[00:11:29]This is divided by two minus one.

[00:11:33]This should be equal to Now, 2 to the power of 1 + square root of 5 to base 2 This is equal to 1 + square root of 5 divided by 2.

[00:11:47]Now, we have 1 + square root of 5 to the power of 2. This is in the form of a + b raised to the power of 2, which we can express as a to the power of 2 + b to the power of 2 + 2 * a b. Now, let's apply this algebraic identity so that we have This is 1 squared.

[00:12:08]Then, we have + square root of 5 raised to the power of 2 then + We have 2 multiplying by square root of 5 divided by 2 subtract 1 This is equal to 1 + square root of 5 to base divided by 2.

[00:12:30]So, this should be equal to Now, from here we have 1 Let's eliminate the square root sign here. So, we have 1 + 5 + 2 square root of 5 divided by 2 - 1 This should give us a value of 1 + square root of 5 over over 2.

[00:12:54]So, here 1 is a whole number, so this is over 1.

[00:12:58]We have 1 + 5, this is 6 + 2 square root of 5 divided by 2 - 1. 1 is a whole number, so this is over 1.

[00:13:10]This should be equal to 1 + square root of 5 over over 2.

[00:13:19]Now, the LCM is 2. Let's multiply everything here by 2 so that 2 and 2 here simplifies.

[00:13:27]Again, we have two and two here simplifies so that we have six plus two square root of five minus one times two, this is two.

[00:13:40]This should be equal to Here we have one plus square root of five.

[00:13:48]Now, we have that two to the power of log one plus square root of five to the power of two to base two minus two.

[00:13:54]And also two to the power of log one plus square root of five to base two minus one. This is in the form of a to the power of n minus m which we can express as a to the power of n over a to the power of m.

[00:14:09]Applying this exponent property, then we have two raised to the power of log one plus square root of five raised to the power of two to base two divided by this is two raised to the power of two, then subtract one. This is equal to two raised to the power of log one plus square root of five to base two divided by two to the power of one.

[00:14:43]Now, we have that two to the power of log one plus square root of five to the power of two to base two and two to the power of log one plus square root of five to base two. This is in the form of a to the power of log b to base a and this should give us a value of b.

[00:14:59]Applying this logarithm property gives for us that two to the power of log one plus square root of five to the power of two to base two, this becomes one plus square root of five squared everything here raised to the power of two.

[00:15:15]Divided by two squared which is four minus one. This is supposed to be equal to Now, two to the power of one plus square root of five to base two, this becomes one plus square root of five divided by two.

[00:15:37]Now, since we have one is a whole number here, the LCM here is four. So, let's multiply everything here by four.

[00:15:46]Here we have times four.

[00:15:50]So, let's simplify four and four here so that we have one plus square root of five raised to the power of two subtract one times four, which is minus four.

[00:16:02]This is supposed to be equal to Now, four divided by two, this is two. So, we have two multiplying by one plus square root of five.

[00:16:14]The next step is Let us expand this equation here. Here we have one squared plus square root of five raised to the power of two.

[00:16:24]Then we have plus This is two times one times square root of five minus four.

[00:16:31]This is supposed to be equal to two times one, which is two plus two times two square root of five.

[00:16:41]Now, here we have one squared is one.

[00:16:44]Let's eliminate the square root sign here so that we have plus five.

[00:16:50]Then plus Now, two times one times square root of five, this is two square root of five.

[00:16:57]Subtract four.

[00:16:59]This should be equal to two plus two square root of five.

[00:17:06]Now, five plus one, this is six. We have six plus two square root of five subtract four.

[00:17:16]This should be to 2 + 2 square root of 5.

[00:17:23]Now, we have 6 - 4. This is 2 + 2 square root of 5.

[00:17:30]And this is equal to 2 + 2 square root of 5.

[00:17:37]So, this indicates that the left and side is equal to the right and side. And this affirms that the value of X the value of X, which is row 1 + square root of 5 to base 2 - 1 satisfies the equation.

[00:17:57]So, kindly follow the steps.

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