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

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

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

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

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

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

Decimal Notation

x = 0.64359 \dots, x = - 0.64359 \dots

Solution steps

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

Apply log rules

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

Solve

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

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

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

lo g_{3} (3 x + 3) = lo g_{3} (12)

0.59 = ln (x)

0 = \frac{100}{ln ( 5 )} ln (x) - 100

ln (x) - \frac{1}{2} x + 1 = 0

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