What is the answer to solvefor y,4ln(y+4)=2e^x+c ?

The answer to solvefor y,4ln(y+4)=2e^x+c is y=e^{(2e^x+c)/4}-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,4ln(y+4)=2e^x+c

solve for

y, 4 ln (y + 4) = 2 e^{x} + c

y = e^{\frac{2 e ^{x} + c}{4}} - 4

Solution steps

4 ln (y + 4) = 2 e^{x} + c

Divide both sides by

4

ln (y + 4) = \frac{2 e ^{x} + c}{4}

Apply log rules

y + 4 = e^{\frac{2 e ^{x} + c}{4}}

Solve

y + 4 = e^{\frac{2 e ^{x} + c}{4}} : y = e^{\frac{2 e ^{x} + c}{4}} - 4

y = e^{\frac{2 e ^{x} + c}{4}} - 4

ln (x) = ln (1) - 1.34

a^{l o g_{a} (x)} = x

lo g_{10} (x + 6) = lo g_{10} (2 x - 1)

\frac{1}{2} ln (x) + c = 0

lo g_{4} (\frac{2}{x + 3}) = - 1

What is the answer to solvefor y,4ln(y+4)=2e^x+c ?
The answer to solvefor y,4ln(y+4)=2e^x+c is y=e^{(2e^x+c)/4}-4