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}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}) = l v

Move

l v

to the left side

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

Factor

v ln (1 - v^{2}) - l v : 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}

- 3 x + 9 = lo g_{2} (x)

- 0.6 = lo g_{10} (x)

lo g_{7} (x + 7) - lo g_{7} (x - 5) = 1

10 \cdot lo g_{10} (x) = 22.9

10 lo g_{10} (x) = 5

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}