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

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

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

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

x = 1, x = - 1

Solution steps

(lo g_{4} (1 + x^{2}))^{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_{5} (x + 3) = 1

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

ln (x - 1) = 0

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

lo g_{10} (x) = 0.5

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