To what power should you raise the squareroot of a number to get the number back. the answer is 2. So sqrt(x) must be 2,.i.e x = 4. The only exception is a number.that if either in the base or the power, the result will be the same as the base no matter what the counterpart is, i.e the number 1. So all we have to do is make sure the either the base or the power, which in this case is the same, is equal to 1. The only number whose squareroot is 1 is the number 1 itself. In other words, you either have to make sure that X is unaffected by the calculation, or that the root of X is countered by the exponent. And the there is one other answer depending on definition. If you go by the approach that 0 to any non-negative power is equal to 0, 0 is also an solution ot this equation
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Let √x = a; => a^2=a^a ; 2loga=aloga; =>2loga - aloga=0;=>loga(2-a)=0;
loga=0 or 2-a=0; =>
a=1;or a=2;recall √x=a;
x=a^2= 1; or 4
By inspection: 1, 4.
To what power should you raise the squareroot of a number to get the number back. the answer is 2. So sqrt(x) must be 2,.i.e x = 4.
The only exception is a number.that if either in the base or the power, the result will be the same as the base no matter what the counterpart is, i.e the number 1. So all we have to do is make sure the either the base or the power, which in this case is the same, is equal to 1. The only number whose squareroot is 1 is the number 1 itself.
In other words, you either have to make sure that X is unaffected by the calculation, or that the root of X is countered by the exponent.
And the there is one other answer depending on definition.
If you go by the approach that 0 to any non-negative power is equal to 0, 0 is also an solution ot this equation
Thanks for your in-depth look into the problem 🥰✅💪✅🙏
(Vx)^2=(Vx)^(Vx) , 2*lnVx=Vx*lnVx , 2=Vx , x=4 , test , 4=(V4)^(V4) , 4=2^2 , 4=4 , OK ,
You obviously have too much paper.