Olympiad Mathematics | Indians 

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

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

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!

8ˣ = x⁶ x = ⁸log(x⁶) x = (⁶⁄₃) • ²log(x) *x = 2 • ²log(x)* *If (x = 2), so* 2 = 2 • ²log(2) 2 = 2 • 1 *2 = 2* ............ _(match)_ *If (x = 4), so* 4 = 2 • ²log(4) 4 = 2 • 2 *4 = 4* ............ _(match)_

x = 2 by inspection. Other solutions might be possible.

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

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.

Why not take log both sides?🤔

X= 2 or 4

The intuitive answer is 2…takes 30 seconds or less. This solution is like using a sledgehammer to crack a walnut

Multiplying!

X= 2.

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It's easy I X=2

Is this olympiad ?

2^3x = x^(3*2) => x=2. math olympiad my аss. stop the bs

2^x = x^2 nhanh hơn

Xis 2

X= 2.