Finite Math Examples

$z = \ln (y - 4 x + 1)$

Step 1

Rewrite the equation as $\ln (y - 4 x + 1) = z$ .

$\ln (y - 4 x + 1) = z$

Step 2

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (y - 4 x + 1)} = e^{z}$

Step 3

Rewrite $\ln (y - 4 x + 1) = z$ 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 $b^{y} = x$ .

$e^{z} = y - 4 x + 1$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $y - 4 x + 1 = e^{z}$ .

$y - 4 x + 1 = e^{z}$

Step 4.2

Move all terms not containing $x$ to the right side of the equation.

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Step 4.2.1

Subtract $y$ from both sides of the equation.

$- 4 x + 1 = e^{z} - y$

Step 4.2.2

Subtract $1$ from both sides of the equation.

$- 4 x = e^{z} - y - 1$

Step 4.3

Divide each term in $- 4 x = e^{z} - y - 1$ by $- 4$ and simplify.

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Step 4.3.1

Divide each term in $- 4 x = e^{z} - y - 1$ by $- 4$ .

$\frac{- 4 x}{- 4} = \frac{e^{z}}{- 4} + \frac{- y}{- 4} + \frac{- 1}{- 4}$

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Cancel the common factor of $- 4$ .

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Step 4.3.2.1.1

Cancel the common factor.

$\frac{- 4 x}{- 4} = \frac{e^{z}}{- 4} + \frac{- y}{- 4} + \frac{- 1}{- 4}$

Step 4.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{e^{z}}{- 4} + \frac{- y}{- 4} + \frac{- 1}{- 4}$

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Simplify each term.

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Step 4.3.3.1.1

Move the negative in front of the fraction.

$x = - \frac{e^{z}}{4} + \frac{- y}{- 4} + \frac{- 1}{- 4}$

Step 4.3.3.1.2

Dividing two negative values results in a positive value.

$x = - \frac{e^{z}}{4} + \frac{y}{4} + \frac{- 1}{- 4}$

Step 4.3.3.1.3

Dividing two negative values results in a positive value.

$x = - \frac{e^{z}}{4} + \frac{y}{4} + \frac{1}{4}$