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=sqrt(\sqrt{4^{0.25)}-1},x=-sqrt(\sqrt{4^{0.25)}-1}

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

The solution is x=sqrt(\sqrt{4^{0.25)}-1},x=-sqrt(\sqrt{4^{0.25)}-1}

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

x = 4^{0.25} - 1, x = - 4^{0.25} - 1

Decimal

x = 0.43497 \dots, x = - 0.43497 \dots

Solution steps

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

Apply log rules

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

Solve

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

x = 4^{0.25} - 1, x = - 4^{0.25} - 1

lo g_{e} (x + 1) + lo g_{e} (x) = 0

lo g_{x - 4} (x) = 3

ln (x) = - ln (x)

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

lo g_{27} (a^{2}) = 18

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=sqrt(\sqrt{4^{0.25)}-1},x=-sqrt(\sqrt{4^{0.25)}-1}
Solve for x: log_{4}((1+x^2)^2)=0.25
The solution is x=sqrt(\sqrt{4^{0.25)}-1},x=-sqrt(\sqrt{4^{0.25)}-1}