who can solve this...?

Question

1^X=2^X
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Answer 1

1^x = 2^x

so 2^x = 1 as 1^x = 1 for all x

so x = 0 as 2^0 = 1 and 2^any number not 0 cannot be 1

so x = 0

Answer 2

x=0

Answer 3

1^x = 2^x

Now, x = x
But 1 does not equal 2.
So, no matter what natural number you put, there is really no way the LHS can equal RHS.

Let's try the only whole number which is not a natural number (I'm talking of 0).

Any number to the power 0 is one.
1^0 = 1
2^0 = 1

Only real value for x that make the equation possible is x = 0.


SECOND METHOD:
1^x is always 1 for any whole number x
1^x = 1 always

So, for the equation to be satisfied, 2^x = 1

This is only possible when x = 0

There are 2 methods. The first one is mixing logic with trial and error. The second is pure logic.

Answer 4

You can do this by inspection.

Since 1 ≠ 2

...then the only solution to 1^x = 2^x is: x = 0

Answer 5

If x = (1 / 1 x 10^10) (or an even greater denominator) then decimally on a calculator you would get that both results to your equation would be 1.

Technically though, x would have to equal (1 / infinity), to make the equation truly equal.

The people above are correct and really the only true answer is that x = 0.

Answer 6

1^X=2^X /: 2^X (notice you have same exponents, so you can play with bases)

1^X / 2^X = 1
(1/2)^X = (1/2)^0 (we used the fact any number with exponent 0 equals 1, and v.v.)

since bases are the same (1/2 in this case), the exponents are equal as well:
x = 0

Check:
1^X=2^X, x = 0
1^0 = 2^0
1 = 1
OK => x = 0

Answer 7

1^x=2^x
As 1^x= 1
2^x = 1
2^0 = 1
Therefore x = 0

Answer 8

x = 0

Any number to the 0 power is 1.

Answer 9

real only for X=0