What is the answer to L(v)=vln(1-v^2) ?

The answer to L(v)=vln(1-v^2) is v=0,v=sqrt(-e^L+1) {L<= 0},v=-sqrt(-e^L+1) {L<= 0}

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L(v)=vln(1-v^2)

L (v) = v ln (1 - v^{2})

v = 0, v = - e^{L} + 1 {L \leq 0}, v = - - e^{L} + 1 {L \leq 0}

Solution steps

L (v) = v ln (1 - v^{2})

Switch sides

v ln (1 - v^{2}) = Lv

Move

Lv

to the left side

v ln (1 - v^{2}) - Lv = 0

Factor

v ln (1 - v^{2}) - Lv : v (ln (1 - v^{2}) - L)

v (ln (1 - v^{2}) - L) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v = 0 or ln (1 - v^{2}) - L = 0

Solve

ln (1 - v^{2}) - L = 0 : v = - e^{L} + 1 {L \leq 0}, v = - - e^{L} + 1 {L \leq 0}

Verify Solutions:

v = 0

True

, v = - e^{L} + 1 {L \leq 0}, v = - - e^{L} + 1 {L \leq 0}

The solutions are

v = 0, v = - e^{L} + 1 {L \leq 0}, v = - - e^{L} + 1 {L \leq 0}

lo g_{b} (81) = 2

lo g_{10} (x^{2} - 16 x + 1) = 0

2 ln (3 x^{2} + 1) = 13

ln (15 - 2 x) = 2 ln (x)

1 = 3 lo g_{10} (4 x + 10)

What is the answer to L(v)=vln(1-v^2) ?
The answer to L(v)=vln(1-v^2) is v=0,v=sqrt(-e^L+1) {L<= 0},v=-sqrt(-e^L+1) {L<= 0}