What is the answer to solvefor y,ln(y-1)-ln(x)=x+ln(x) ?

The answer to solvefor y,ln(y-1)-ln(x)=x+ln(x) is y=e^xx^2+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor y,ln(y-1)-ln(x)=x+ln(x)

solve for

y, ln (y - 1) - ln (x) = x + ln (x)

y = e^{x} x^{2} + 1

Solution steps

ln (y - 1) - ln (x) = x + ln (x)

Move

ln (x)

to the right side

ln (y - 1) = x + 2 ln (x)

Apply log rules

y - 1 = e^{x} x^{2}

Solve

y - 1 = e^{x} x^{2} : y = e^{x} x^{2} + 1

y = e^{x} x^{2} + 1

Verify Solutions:

y = e^{x} x^{2} + 1

True

The solution is

y = e^{x} x^{2} + 1

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

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

lo g_{7} (2 x + 1) = 2

lo g_{a} (216) = 3

lo g_{3} (11 - 4 x) = 3

What is the answer to solvefor y,ln(y-1)-ln(x)=x+ln(x) ?
The answer to solvefor y,ln(y-1)-ln(x)=x+ln(x) is y=e^xx^2+1