What is the answer to (log_{0.5}(1-x^2))^2=4 ?

The answer to (log_{0.5}(1-x^2))^2=4 is x=(sqrt(3))/2 ,x=-(sqrt(3))/2

Solve for x: (log_{0.5}(1-x^2))^2=4

The solution is x=(sqrt(3))/2 ,x=-(sqrt(3))/2

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(lo g_{0.5} (1 - x^{2}))^{2} = 4

x = \frac{3}{2}, x = - \frac{3}{2}

Decimal Notation

x = 0.86602 \dots, x = - 0.86602 \dots

Solution steps

(lo g_{0.5} (1 - x^{2}))^{2} = 4

Rewrite the equation with

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

(u)^{2} = 4

Solve

u^{2} = 4 : u = 4, u = - 4

u = 4, u = - 4

Substitute back

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

solve for

x

Solve

lo g_{0.5} (1 - x^{2}) = 4 : x = \frac{3}{2}, x = - \frac{3}{2}

Solve

lo g_{0.5} (1 - x^{2}) = - 4 :

No Solution for

x \in R

x = \frac{3}{2}, x = - \frac{3}{2}

Verify Solutions:

x = \frac{3}{2}

True

, x = - \frac{3}{2}

True

The solutions are

x = \frac{3}{2}, x = - \frac{3}{2}

lo g_{2} (x - 1) = 4

x, y - π = ln (x)

5000 lo g_{10} (n) = n

lo g_{2} ((r + 18)^{2}) = 6

lo g_{10} (2 x + 1) = lo g_{10} (3 x)

What is the answer to (log_{0.5}(1-x^2))^2=4 ?
The answer to (log_{0.5}(1-x^2))^2=4 is x=(sqrt(3))/2 ,x=-(sqrt(3))/2
Solve for x: (log_{0.5}(1-x^2))^2=4
The solution is x=(sqrt(3))/2 ,x=-(sqrt(3))/2