Another approach: 8ˣ = x⁶ Taking the logarithm of both sides gives xln8 = 6ln|x| Using the power rule of logarithms gives x/ln|x| = 6/ln8 = 6/3ln2 = 2/ln2 and x/ln|x| = 2/ln2 = 4/2ln2 = 4/ln(2²) = 4/ln4 For x > 0, we have x/ln|x| = x/lnx so x/lnx = 2/ln2 and x/lnx = 4/ln4 which means that, for x > 0, there are two solutions: x = 2 and x = 4 Note that 8² = 2⁶ = 64 and 8⁴ = 4⁶ = 2¹² = 4096 For x < 0, we have x/ln|x| = x/ln(-x), so x/ln(-x) = 2/ln2 where, it turns out, there is a third solution: x ≈ -0.7666646959631231
x=2 is only one of three solutions to this problem. 4 and ~-0.766665 are the others. I think you would be expected to demonstrate how to find these in an "Olympiad" question
If we imagine graphs of functions 8^x and x^6, it is clear that there should be at least 2 intersections at x0. Solution x=2. obviously, and you need to find the second one in the negative area. But it turns out there is a third solution!
It may help to consider revisiting some laws and rules in algebra specifically on identities of basic operations, i.e. 'cancellation of terms'. The art of 'rearranging' terms or their components takes practice using different approaches.
Easy to find x= 2 but are there any values for x? If we can prove f(x)=x^(1/x) is a monotonic function, we can say x=2 is only solution. However, this problem has many solutions, example x=4 is another solution
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Another approach:
8ˣ = x⁶
Taking the logarithm of both sides gives
xln8 = 6ln|x|
Using the power rule of logarithms gives
x/ln|x| = 6/ln8 = 6/3ln2 = 2/ln2
and
x/ln|x| = 2/ln2 = 4/2ln2 = 4/ln(2²) = 4/ln4
For x > 0, we have x/ln|x| = x/lnx so
x/lnx = 2/ln2 and x/lnx = 4/ln4
which means that, for x > 0, there are two solutions:
x = 2 and x = 4
Note that 8² = 2⁶ = 64 and 8⁴ = 4⁶ = 2¹² = 4096
For x < 0, we have x/ln|x| = x/ln(-x),
so x/ln(-x) = 2/ln2
where, it turns out, there is a third solution:
x ≈ -0.7666646959631231
x=2 is only one of three solutions to this problem.
4 and ~-0.766665 are the others. I think you would be expected to demonstrate how to find these in an "Olympiad" question
x = 2 by inspection. Other solutions might be possible.
Aap bahut long chale gaye.
If we imagine graphs of functions 8^x and x^6, it is clear that there should be at least 2 intersections at x0.
Solution x=2. obviously, and you need to find the second one in the negative area.
But it turns out there is a third solution!
It may help to consider revisiting some laws and rules in algebra specifically on identities of basic operations, i.e. 'cancellation of terms'. The art of 'rearranging' terms or their components takes practice using different approaches.
The solution is incomplete.
I randomly take 2 and it right. Maybe intuition
❤❤❤❤❤❤❤❤❤❤❤
Easy to find x= 2 but are there any values for x? If we can prove f(x)=x^(1/x) is a monotonic function, we can say x=2 is only solution. However, this problem has many solutions, example x=4 is another solution
Asaranai neda paw.
Multiplying!
It's easy I X=2
2^3x = x^(3*2) => x=2. math olympiad my аss. stop the bs
Why not take log both sides?🤔
X= 2 or 4
X= 2.
Is this olympiad ?
2^x = x^2 nhanh hơn
Xis 2
The intuitive answer is 2…takes 30 seconds or less. This solution is like using a sledgehammer to crack a walnut
X= 2.