What is the answer to 2x=ln(3+x) ?

The answer to 2x=ln(3+x) is x=-(\W_{-1}(-2/(e^6))+6)/2 ,x=-(\W_{0}(-2/(e^6))+6)/2

Solve for x: 2x=ln(3+x)

The solution is x=-(\W_{-1}(-2/(e^6))+6)/2 ,x=-(\W_{0}(-2/(e^6))+6)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x = ln (3 + x)

x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

Decimal Notation

x = 0.64694 \dots, x = - 2.99750 \dots

Solution steps

2 x = ln (3 + x)

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{2 x} = e^{l n (3 + x)}

Simplify

e^{l n (3 + x)} : 3 + x

e^{2 x} = 3 + x

Solve

e^{2 x} = 3 + x : x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

Verify Solutions:

x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}

True

, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

True

The solutions are

x = - \frac{W _{- 1} ( - \frac{2}{e ^{6}} ) + 6}{2}, x = - \frac{W _{0} ( - \frac{2}{e ^{6}} ) + 6}{2}

4 ln (x) = ln (4 x)

lo g_{3} (x + 9) + lo g_{3} (x + 1) = 1

x \frac{1}{240} = ln (x)

ln^{(2)} (x^{(3)}) = e

e^{- 2 l n (x)} = - 2 x

What is the answer to 2x=ln(3+x) ?
The answer to 2x=ln(3+x) is x=-(\W_{-1}(-2/(e^6))+6)/2 ,x=-(\W_{0}(-2/(e^6))+6)/2
Solve for x: 2x=ln(3+x)
The solution is x=-(\W_{-1}(-2/(e^6))+6)/2 ,x=-(\W_{0}(-2/(e^6))+6)/2