What is the answer to solvefor y,x=sqrt(1-log_{0.5)(2-y)} ?

The answer to solvefor y,x=sqrt(1-log_{0.5)(2-y)} is y=-0.5^{1-x^2}+2 {x>= 0}

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solvefor y,x=sqrt(1-log_{0.5)(2-y)}

solve for

y, x = 1 - lo g_{0.5} (2 - y)

y = - 0. 5^{1 - x^{2}} + 2 {x \geq 0}

Solution steps

x = 1 - lo g_{0.5} (2 - y)

Rewrite the equation with

lo g_{0.5} (2 - y) = u

x = 1 - u

Solve

x = 1 - u : u = 1 - x^{2} {x \geq 0}

u = 1 - x^{2} {x \geq 0}

Substitute back

u = lo g_{0.5} (2 - y),

solve for

y

Solve

lo g_{0.5} (2 - y) = 1 - x^{2} : y = - 0. 5^{1 - x^{2}} + 2

y = - 0. 5^{1 - x^{2}} + 2

Verify Solutions:

y = - 0. 5^{1 - x^{2}} + 2 {x \geq 0}

The solution is

y = - 0. 5^{1 - x^{2}} + 2 {x \geq 0}

7 = lo g_{10} (\frac{I}{1})

ln (x) + ln (x + 2) = 3

lo g_{2} (2 x + 1) = 6

- 2 \cdot ln (2 w - 3) = - 10

ln (x) + ln (x - 10) = 2

What is the answer to solvefor y,x=sqrt(1-log_{0.5)(2-y)} ?
The answer to solvefor y,x=sqrt(1-log_{0.5)(2-y)} is y=-0.5^{1-x^2}+2 {x>= 0}