Finite Math Examples

$2^{2 x} - 3^{2 y} = 55$

Step 1

Solve the equation for $y$ .

Tap for more steps...

Step 1.1

Subtract $2^{2 x}$ from both sides of the equation.

$- 3^{2 y} = 55 - 2^{2 x}$

Step 1.2

Divide each term in $- 3^{2 y} = 55 - 2^{2 x}$ by $- 1$ and simplify.

Tap for more steps...

Step 1.2.1

Divide each term in $- 3^{2 y} = 55 - 2^{2 x}$ by $- 1$ .

$\frac{- 3^{2 y}}{- 1} = \frac{55}{- 1} + \frac{- 2^{2 x}}{- 1}$

Step 1.2.2

Simplify the left side.

Tap for more steps...

Step 1.2.2.1

Dividing two negative values results in a positive value.

$\frac{3^{2 y}}{1} = \frac{55}{- 1} + \frac{- 2^{2 x}}{- 1}$

Step 1.2.2.2

Divide $3^{2 y}$ by $1$ .

$3^{2 y} = \frac{55}{- 1} + \frac{- 2^{2 x}}{- 1}$

Step 1.2.3

Simplify the right side.

Tap for more steps...

Step 1.2.3.1

Simplify each term.

Tap for more steps...

Step 1.2.3.1.1

Divide $55$ by $- 1$ .

$3^{2 y} = - 55 + \frac{- 2^{2 x}}{- 1}$

Step 1.2.3.1.2

Dividing two negative values results in a positive value.

$3^{2 y} = - 55 + \frac{2^{2 x}}{1}$

Step 1.2.3.1.3

Divide $2^{2 x}$ by $1$ .

$3^{2 y} = - 55 + 2^{2 x}$

Step 1.3

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

$\ln (3^{2 y}) = \ln (- 55 + 2^{2 x})$

Step 1.4

Expand $\ln (3^{2 y})$ by moving $2 y$ outside the logarithm.

$2 y \ln (3) = \ln (- 55 + 2^{2 x})$

Step 1.5

Divide each term in $2 y \ln (3) = \ln (- 55 + 2^{2 x})$ by $2 \ln (3)$ and simplify.

Tap for more steps...

Step 1.5.1

Divide each term in $2 y \ln (3) = \ln (- 55 + 2^{2 x})$ by $2 \ln (3)$ .

$\frac{2 y \ln (3)}{2 \ln (3)} = \frac{\ln (- 55 + 2^{2 x})}{2 \ln (3)}$

Step 1.5.2

Simplify the left side.

Tap for more steps...

Step 1.5.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.5.2.1.1

Cancel the common factor.

$\frac{2 y \ln (3)}{2 \ln (3)} = \frac{\ln (- 55 + 2^{2 x})}{2 \ln (3)}$

Step 1.5.2.1.2

Rewrite the expression.

$\frac{y \ln (3)}{\ln (3)} = \frac{\ln (- 55 + 2^{2 x})}{2 \ln (3)}$

Step 1.5.2.2

Cancel the common factor of $\ln (3)$ .

Tap for more steps...

Step 1.5.2.2.1

Cancel the common factor.

$\frac{y \ln (3)}{\ln (3)} = \frac{\ln (- 55 + 2^{2 x})}{2 \ln (3)}$

Step 1.5.2.2.2

Divide $y$ by $1$ .

$y = \frac{\ln (- 55 + 2^{2 x})}{2 \ln (3)}$

Step 2

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be $0$ or $1$ for each of its variables. In this case, the degree of the variable in the equation violates the linear equation definition, which means that the equation is not a linear equation.

Not Linear