What is the answer to log^2_{4}(1+x^2)=0.24 ?

The answer to log^2_{4}(1+x^2)=0.24 is x=sqrt(4^{\sqrt{0.24)}-1},x=-sqrt(4^{\sqrt{0.24)}-1}

Solve for x: log^2_{4}(1+x^2)=0.24

The solution is x=sqrt(4^{\sqrt{0.24)}-1},x=-sqrt(4^{\sqrt{0.24)}-1}

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lo g_{4}^{2} (1 + x^{2}) = 0.24

x = 4^{0.24} - 1, x = - 4^{0.24} - 1

Decimal Notation

x = 0.98599 \dots, x = - 0.98599 \dots

Solution steps

lo g_{4}^{2} (1 + x^{2}) = 0.24

Rewrite the equation with

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

u^{2} = 0.24

Solve

u^{2} = 0.24 : u = 0.24, u = - 0.24

u = 0.24, u = - 0.24

Substitute back

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

solve for

x

Solve

lo g_{4} (1 + x^{2}) = 0.24 : x = 4^{0.24} - 1, x = - 4^{0.24} - 1

Solve

lo g_{4} (1 + x^{2}) = - 0.24 :

No Solution for

x \in R

x = 4^{0.24} - 1, x = - 4^{0.24} - 1

Verify Solutions:

x = 4^{0.24} - 1

True

, x = - 4^{0.24} - 1

True

The solutions are

x = 4^{0.24} - 1, x = - 4^{0.24} - 1

lo g_{2} (x^{2} - 1) = 1 0^{10}

ln (x) = - \frac{1}{20}

lo g_{10} (x) = - 0.33

x, lo g_{2} (x + y) = 3

lo g_{10} (a) = 0.19

What is the answer to log^2_{4}(1+x^2)=0.24 ?
The answer to log^2_{4}(1+x^2)=0.24 is x=sqrt(4^{\sqrt{0.24)}-1},x=-sqrt(4^{\sqrt{0.24)}-1}
Solve for x: log^2_{4}(1+x^2)=0.24
The solution is x=sqrt(4^{\sqrt{0.24)}-1},x=-sqrt(4^{\sqrt{0.24)}-1}