What is the answer to -ln(4-z^2)+c=0 ?

The answer to -ln(4-z^2)+c=0 is z=sqrt(-e^c+4) {c<= 2ln(2)},z=-sqrt(-e^c+4) {c<= 2ln(2)}

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-ln(4-z^2)+c=0

- ln (4 - z^{2}) + c = 0

z = - e^{c} + 4 {c \leq 2 ln (2)}, z = - - e^{c} + 4 {c \leq 2 ln (2)}

Solution steps

- ln (4 - z^{2}) + c = 0

Move

c

to the right side

- ln (4 - z^{2}) = - c

Divide both sides by

- 1

ln (4 - z^{2}) = c

Apply log rules

4 - z^{2} = e^{c}

Solve

4 - z^{2} = e^{c} : z = - e^{c} + 4, z = - - e^{c} + 4

z = - e^{c} + 4, z = - - e^{c} + 4

Verify Solutions:

z = - e^{c} + 4 {c \leq 2 ln (2)}, z = - - e^{c} + 4 {c \leq 2 ln (2)}

The solutions are

z = - e^{c} + 4 {c \leq 2 ln (2)}, z = - - e^{c} + 4 {c \leq 2 ln (2)}

x = lo g_{10} (3 (5^{x}))

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

x, 0.41 = lo g_{10} (x)

x = ln (19 t), t

ln (x) = \frac{1}{5}

What is the answer to -ln(4-z^2)+c=0 ?
The answer to -ln(4-z^2)+c=0 is z=sqrt(-e^c+4) {c<= 2ln(2)},z=-sqrt(-e^c+4) {c<= 2ln(2)}