What is the answer to log^2_{4}(1+x^2)=0.25 ?

The answer to log^2_{4}(1+x^2)=0.25 is x=1,x=-1

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log^2_{4}(1+x^2)=0.25

lo g_{4}^{2} (1 + x^{2}) = 0.25

x = 1, x = - 1

Solution steps

lo g_{4}^{2} (1 + x^{2}) = 0.25

Rewrite the equation with

lo g_{4} (1 + x^{2}) = u

u^{2} = 0.25

Solve

u^{2} = 0.25 : u = 0.25, u = - 0.25

u = 0.25, u = - 0.25

Substitute back

u = lo g_{4} (1 + x^{2}),

solve for

x

Solve

lo g_{4} (1 + x^{2}) = 0.25 : x = 1, x = - 1

Solve

lo g_{4} (1 + x^{2}) = - 0.25 :

No Solution for

x \in R

x = 1, x = - 1

Verify Solutions:

x = 1

True

, x = - 1

True

The solutions are

x = 1, x = - 1

lo g_{10} (\frac{x}{3}) = 2

lo g_{x} (16) = 4

ln (ln (x)) = 0

lo g_{4} (a) = lo g_{2} (x)

lo g_{3} (x + 3) = lo g_{3} (2 x - 5)

What is the answer to log^2_{4}(1+x^2)=0.25 ?
The answer to log^2_{4}(1+x^2)=0.25 is x=1,x=-1
Solve for x: log^2_{4}(1+x^2)=0.25
The solution is x=1,x=-1