6^x=6x+24 Divide by 6: 6^(x-1)=x+4 Note that x>1 and an integer. Thus x is an integer greater or equal to 2. Take modulo 2: [mod(6,2)]^(x-1)=mod(x,2) mod(x,2)=0 --> x is even If x=2, LHS=6^x =6² --> LHS =36 RHS=6x+24 =6(2)+24 --> RHS=36 For an even integer greater than 2 LHS increase drastically but not RHS. Therefore x=2.
By hit and trial method. Let’s substitute x = 1, we have 6^1 = 6 and 6*1 + 24 = 30 6 = 30 , which is not correct try substituting x = 2 , we have 6^2 = 36, also 6*2 + 24 = 36 Therefore solution for x is equal to 2
The answer is x=2. I was just wondering how similar is the thought process compared to solving for x in 2^x+x=5???Or what differences??? I am trying to spot any differences so that when I practice all Lambert W problems, I easily follow on how the rules of algebra are "transcended". I also hope that you do have a playlist of Lambert W problems because if you do I would like to compare and contrast that with *other* playlists.
Lambert W problems are just similarity problems written in a more convoluted form. e.g. 7^(x-3) / (x-3) = 7^5 / 5 --> (x-3) = 5 --> x = 8. The trick is in recognising how to get the equation into a form where the similarity equality applies. Lambert W dies that fir the limited case of log(x) related to x relationships. Similarity extends further to other functions and classes of functions.
If you let f(x) = 6ˣ - (6x + 24) = 0 By observation, x = 2 is a solution. As 6ˣ grows much faster than 6x + 24 for all x > 2, then there are no more solutions to f(x) for x > 2. The other real solution is x = -3.999871 (approximately) which can be found using the Lambert W function or approximately to any degree of significance using the Newton-Raphson method.
@@michaeldoerr5810 Possibly - if you can find the right conditions to set up the similarity. But similarity with real numbers isn't assured. And some equations result in absurdities that have no solution of any type. The problem you pose appears at first blush to have no real solutions. At x=3 7^3 is 0 wrong. At x=4 the LHS is increasing exponentially compared to the RHS, so no positive real solution x>=3 exists. Any values less than three result in LHS becoming increasing small and then inverts, while the RGS is negative. No real solutions. Etc... (click to see the rest) As with many of these problems, analytic solutions rely on the problem being setup to allow analytic solutions. So e.g. Start with 7^(x+5) / (x+5) = 7^7 / 7, so x = 2. now rewrite the equation as 7^(x+5) = 7^6 * (x+5). then 7^x = (7^6 / 7^5) * (x+5). then 7^x = 7x + 35. That is now the problem to solve. Change any of the values and the equation can become quite difficult to solve analytically, or simply impossible. Many test questions are created in this way. If you recognize the tricks or transformations, reversing them is straight forward. In other cases, estimating solutions by examining the behavior of the components or rearranged versions allows partial solutions to pop out allowing the question to be simplified, or narrowing down the range of the solutions. The key is that such questions have answers. Knowing they have answers allows solving them. Were I designing the exams, I would include traps where the equations are impossible, or where disallowed values result unless you see the rearrangement that makes clear that they are disallowed. In the real world of applied math, such things happen. Recognising them is vital. Failing to recognize them can lead to very expensive equipment tearing itself apart as it trues to reach those impossible or imaginary conditions. People can be hurt or killed. This extends beyond the maths too. In older times relays were used to drive equipment. Relay logic was used to control systems. Programming relay logic is an art. Things like "rely races" occur, where the solution of the system state can become ambiguous or impossible. And again, bad outcomes result. Those relay logic systems were replaced by programmable logic computers that implemented relay logic as software. That resolved some problems and created new ones. etc...
Harvard University Admission Exam Tricks: 6ˣ = 6x + 24; x =? x ≠ 0; x > 1, 24 > 6ˣ or x < - 1, 6ˣ >/≈ 0 x > 1: 6ˣ - 6x = 24 = 6(4) = 6(6 - 2) = 6² - 12 = 6² - 6(2); x = 2 x < - 1, x = - 4: 6⁻⁴ - 6(- 4) = 0.0008 + 24 ≈ 24; x = - 4: 6⁻⁴ - 6(- 4) > 24; Proved The calculation was achieved on a smartphone with a standard calculator app Answer check: x = 2: 6ˣ = 6x + 24; Confirmed as shown x = - 4: 6ˣ = 6x + 24; Confirmed as shown Final answer: x = 2 or x = - 4
6^x=6x+24
Divide by 6: 6^(x-1)=x+4
Note that x>1 and an integer. Thus x is an integer greater or equal to 2.
Take modulo 2:
[mod(6,2)]^(x-1)=mod(x,2)
mod(x,2)=0 --> x is even
If x=2, LHS=6^x
=6² --> LHS =36
RHS=6x+24
=6(2)+24 --> RHS=36
For an even integer greater than 2 LHS increase drastically but not RHS. Therefore x=2.
By hit and trial method.
Let’s substitute x = 1, we have
6^1 = 6 and
6*1 + 24 = 30
6 = 30 , which is not correct
try substituting x = 2 , we have
6^2 = 36, also
6*2 + 24 = 36
Therefore solution for x is equal to 2
There's second solution by Lambert W function
Piece of cake. 6^x = 6x +24 --> 6^(x+4) = 6^4 * (6x +24) --> 6^(x+4) = 6^5 * (x+4) --> 6^(x+4) / (x+4) = 6^5 --> 6^(x+4) / (x+4) = 6^6 / 6 ==> (x+4) = 6 --> x=2. QED
6^х-6х=24 6^xln6-6=0 x=log6(6/ln6) -*+>x min=6log6(e/6*ln6)
6^x-6*x=24
6^x-6*x=6^2-2*6
Thus , x=2
6x̌=6×+24. /6×6=36. 6×2=12
36=12+24=36
×=2
6ˣ⁻¹ = x + 4
x + 4 = u => x = u - 4
6ᵘ⁻⁵ = u
u6⁻ᵘ = 6⁻⁵
-u6⁻ᵘ = -6⁻⁵
-u6⁻ᵘ = (-6)6⁻⁶
-u = -6 => u = 6
x = u - 4 => *x = 2*
This is very powerful and resourceful. Which means that you obtain solutions by comparison. Lambert W function is NOT necessary for the 1st solution 👌
@@superacademy247 yes. If we are searching for only one solution (any possible solution) then W Lambert function is not necessary
🤌
By option , putting 1,2,3.... So putting 2 , we can find 36=36
The answer is x=2. I was just wondering how similar is the thought process compared to solving for x in 2^x+x=5???Or what differences??? I am trying to spot any differences so that when I practice all Lambert W problems, I easily follow on how the rules of algebra are "transcended". I also hope that you do have a playlist of Lambert W problems because if you do I would like to compare and contrast that with *other* playlists.
Lambert W problems are just similarity problems written in a more convoluted form. e.g. 7^(x-3) / (x-3) = 7^5 / 5 --> (x-3) = 5 --> x = 8. The trick is in recognising how to get the equation into a form where the similarity equality applies. Lambert W dies that fir the limited case of log(x) related to x relationships. Similarity extends further to other functions and classes of functions.
@@tunneloflight That similarity problem that you posted this is similar to 7^x=7x-21. Correct???
I'll be posting similar question to help you grasp key concepts of Lambert W function
If you let f(x) = 6ˣ - (6x + 24) = 0
By observation, x = 2 is a solution.
As 6ˣ grows much faster than 6x + 24 for all x > 2, then there are no more solutions to f(x) for x > 2.
The other real solution is x = -3.999871 (approximately) which can be found using the Lambert W function or approximately to any degree of significance using the Newton-Raphson method.
@@michaeldoerr5810 Possibly - if you can find the right conditions to set up the similarity. But similarity with real numbers isn't assured. And some equations result in absurdities that have no solution of any type. The problem you pose appears at first blush to have no real solutions. At x=3 7^3 is 0 wrong. At x=4 the LHS is increasing exponentially compared to the RHS, so no positive real solution x>=3 exists. Any values less than three result in LHS becoming increasing small and then inverts, while the RGS is negative. No real solutions. Etc... (click to see the rest)
As with many of these problems, analytic solutions rely on the problem being setup to allow analytic solutions. So e.g. Start with 7^(x+5) / (x+5) = 7^7 / 7, so x = 2. now rewrite the equation as 7^(x+5) = 7^6 * (x+5). then 7^x = (7^6 / 7^5) * (x+5). then 7^x = 7x + 35. That is now the problem to solve. Change any of the values and the equation can become quite difficult to solve analytically, or simply impossible. Many test questions are created in this way. If you recognize the tricks or transformations, reversing them is straight forward.
In other cases, estimating solutions by examining the behavior of the components or rearranged versions allows partial solutions to pop out allowing the question to be simplified, or narrowing down the range of the solutions.
The key is that such questions have answers. Knowing they have answers allows solving them. Were I designing the exams, I would include traps where the equations are impossible, or where disallowed values result unless you see the rearrangement that makes clear that they are disallowed.
In the real world of applied math, such things happen. Recognising them is vital. Failing to recognize them can lead to very expensive equipment tearing itself apart as it trues to reach those impossible or imaginary conditions. People can be hurt or killed.
This extends beyond the maths too. In older times relays were used to drive equipment. Relay logic was used to control systems. Programming relay logic is an art. Things like "rely races" occur, where the solution of the system state can become ambiguous or impossible. And again, bad outcomes result.
Those relay logic systems were replaced by programmable logic computers that implemented relay logic as software. That resolved some problems and created new ones. etc...
Am i the only one who could guess the answer as x=2 after looking at the question 😢?
Harvard University Admission Exam Tricks: 6ˣ = 6x + 24; x =?
x ≠ 0; x > 1, 24 > 6ˣ or x < - 1, 6ˣ >/≈ 0
x > 1: 6ˣ - 6x = 24 = 6(4) = 6(6 - 2) = 6² - 12 = 6² - 6(2); x = 2
x < - 1, x = - 4: 6⁻⁴ - 6(- 4) = 0.0008 + 24 ≈ 24; x = - 4: 6⁻⁴ - 6(- 4) > 24; Proved
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
x = 2: 6ˣ = 6x + 24; Confirmed as shown
x = - 4: 6ˣ = 6x + 24; Confirmed as shown
Final answer:
x = 2 or x = - 4
One thing, the other solution you show it graphically ok nice but you need to prove it.
Graphical solutions are self explanatory! Calculations without graphs require proofs.
no sir you are wrong , if you want an exact answer you have to prove it ,the second solution is clearly irrational @@superacademy247
By guessing, we can easily find x=2 but then we would need the Lambert W function to find all solutions.
Hit n trial be like: finish in 5 seconds
But you must demonstrate authority over the topic by showing all steps.
I dont think that you need to give too much time to explain the basics ..when you do this level of math equations xD
6^x growing faster so x=2 that's it