What are the solutions to the equation ln(4^3+4)=x^2+ln(2) ?

The solutions to the equation ln(4^3+4)=x^2+ln(2) are x=sqrt(ln(34)),x=-sqrt(ln(34))

Find the zeros of ln(4^3+4)=x^2+ln(2)

The zeros of ln(4^3+4)=x^2+ln(2) are x=sqrt(ln(34)),x=-sqrt(ln(34))

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ln (4^{3} + 4) = x^{2} + ln (2)

x = ln (34), x = - ln (34)

Decimal Notation

x = 1.87786 \dots, x = - 1.87786 \dots

Solution steps

ln (4^{3} + 4) = x^{2} + ln (2)

Switch sides

x^{2} + ln (2) = ln (4^{3} + 4)

Move

ln (2)

to the right side

x^{2} = ln (34)

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = ln (34), x = - ln (34)

x^{3} + x^{2} - 32 x - 60

4 \leq x \leq 8

z^{3} - 27

x = 5 - 7

3 = - 2 (- \frac{5}{4} x + \frac{7}{8}) + x

What are the solutions to the equation ln(4^3+4)=x^2+ln(2) ?
The solutions to the equation ln(4^3+4)=x^2+ln(2) are x=sqrt(ln(34)),x=-sqrt(ln(34))
Find the zeros of ln(4^3+4)=x^2+ln(2)
The zeros of ln(4^3+4)=x^2+ln(2) are x=sqrt(ln(34)),x=-sqrt(ln(34))