Pre-Algebra Examples

x + ln (y) - x^{2} y^{3} = 0

$x + \ln (y) - x^{2} y^{3} = 0$

Step 1

Rewrite in slope-intercept form.

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Step 1.1

The slope-intercept form is

$y = m x + b$ , where

$m$ is the slope and

$b$ is the y-intercept.

$y = m x + b$

Step 1.2

To solve for

$y$ , rewrite the equation using properties of logarithms.

$e^{\ln (y)} = e^{- x + x^{2} y^{3}}$

Step 1.3

Rewrite

$\ln (y) = - x + x^{2} y^{3}$ in exponential form using the definition of a logarithm. If

$x$ and

$b$ are positive real numbers and

$b \neq 1$ , then

$\log_{b} (x) = y$ is equivalent to

$b^{y} = x$ .

$e^{- x + x^{2} y^{3}} = y$

Step 1.4

Solve for

$y$ .

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Step 1.4.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- x + x^{2} y^{3}}) = \ln (y)$

Step 1.4.2

Expand the left side.

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Step 1.4.2.1

Expand

$\ln (e^{- x + x^{2} y^{3}})$ by moving

$- x + x^{2} y^{3}$ outside the logarithm.

$(- x + x^{2} y^{3}) \ln (e) = \ln (y)$

Step 1.4.2.2

The natural logarithm of

$e$ is

$1$ .

$(- x + x^{2} y^{3}) \cdot 1 = \ln (y)$

Step 1.4.2.3

Multiply

$- x + x^{2} y^{3}$ by

$1$ .

$- x + x^{2} y^{3} = \ln (y)$

Step 1.4.3

Subtract

$\ln (y)$ from both sides of the equation.

$- x + x^{2} y^{3} - \ln (y) = 0$

Step 1.4.4

To solve for

$y$ , rewrite the equation using properties of logarithms.

$e^{\ln (y)} = e^{- x + x^{2} y^{3}}$

Step 1.4.5

Rewrite

$\ln (y) = - x + x^{2} y^{3}$ in exponential form using the definition of a logarithm. If

$x$ and

$b$ are positive real numbers and

$b \neq 1$ , then

$\log_{b} (x) = y$ is equivalent to

$b^{y} = x$ .

$e^{- x + x^{2} y^{3}} = y$

Step 1.4.6

Solve for

$y$ .

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Step 1.4.6.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- x + x^{2} y^{3}}) = \ln (y)$

Step 1.4.6.2

Expand the left side.

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Step 1.4.6.2.1

Expand

$\ln (e^{- x + x^{2} y^{3}})$ by moving

$- x + x^{2} y^{3}$ outside the logarithm.

$(- x + x^{2} y^{3}) \ln (e) = \ln (y)$

Step 1.4.6.2.2

The natural logarithm of

$e$ is

$1$ .

$(- x + x^{2} y^{3}) \cdot 1 = \ln (y)$

Step 1.4.6.2.3

Multiply

$- x + x^{2} y^{3}$ by

$1$ .

$- x + x^{2} y^{3} = \ln (y)$

Step 1.4.6.3

Subtract

$\ln (y)$ from both sides of the equation.

$- x + x^{2} y^{3} - \ln (y) = 0$

Step 1.4.6.4

To solve for

$y$ , rewrite the equation using properties of logarithms.

$e^{\ln (y)} = e^{- x + x^{2} y^{3}}$

Step 1.4.6.5

Rewrite

$\ln (y) = - x + x^{2} y^{3}$ in exponential form using the definition of a logarithm. If

$x$ and

$b$ are positive real numbers and

$b \neq 1$ , then

$\log_{b} (x) = y$ is equivalent to

$b^{y} = x$ .

$e^{- x + x^{2} y^{3}} = y$

Step 1.4.6.6

Solve for

$y$ .

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Step 1.4.6.6.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- x + x^{2} y^{3}}) = \ln (y)$

Step 1.4.6.6.2

Expand the left side.

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Step 1.4.6.6.2.1

Expand

$\ln (e^{- x + x^{2} y^{3}})$ by moving

$- x + x^{2} y^{3}$ outside the logarithm.

$(- x + x^{2} y^{3}) \ln (e) = \ln (y)$

Step 1.4.6.6.2.2

The natural logarithm of

$e$ is

$1$ .

$(- x + x^{2} y^{3}) \cdot 1 = \ln (y)$

Step 1.4.6.6.2.3

Multiply

$- x + x^{2} y^{3}$ by

$1$ .

$- x + x^{2} y^{3} = \ln (y)$

Step 1.4.6.6.3

Subtract

$\ln (y)$ from both sides of the equation.

$- x + x^{2} y^{3} - \ln (y) = 0$

Step 1.4.6.6.4

To solve for

$y$ , rewrite the equation using properties of logarithms.

$e^{\ln (y)} = e^{- x + x^{2} y^{3}}$

Step 1.4.6.6.5

Rewrite

$\ln (y) = - x + x^{2} y^{3}$ in exponential form using the definition of a logarithm. If

$x$ and

$b$ are positive real numbers and

$b \neq 1$ , then

$\log_{b} (x) = y$ is equivalent to

$b^{y} = x$ .

$e^{- x + x^{2} y^{3}} = y$

Step 1.4.6.6.6

Solve for

$y$ .

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Step 1.4.6.6.6.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- x + x^{2} y^{3}}) = \ln (y)$

Step 1.4.6.6.6.2

Expand the left side.

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Step 1.4.6.6.6.2.1

Expand

$\ln (e^{- x + x^{2} y^{3}})$ by moving

$- x + x^{2} y^{3}$ outside the logarithm.

$(- x + x^{2} y^{3}) \ln (e) = \ln (y)$

Step 1.4.6.6.6.2.2

The natural logarithm of

$e$ is

$1$ .

$(- x + x^{2} y^{3}) \cdot 1 = \ln (y)$

Step 1.4.6.6.6.2.3

Multiply

$- x + x^{2} y^{3}$ by

$1$ .

$- x + x^{2} y^{3} = \ln (y)$

Step 2

The equation is not linear, so a constant slope does not exist.

Not Linear