Finite Math Examples

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

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

Step 1

Solve the equation for

y

$y$ .

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Step 1.1

Subtract

2^{2 x}

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

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

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

Step 1.2

Divide each term in

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

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

- 1

$- 1$ and simplify.

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Step 1.2.1

Divide each term in

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

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

- 1

$- 1$ .

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

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

Step 1.2.2

Simplify the left side.

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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}

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

Step 1.2.2.2

Divide

3^{2 y}

$3^{2 y}$ by

1

$1$ .

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

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

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

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

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide

55

$55$ by

- 1

$- 1$ .

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

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

Step 1.2.3.1.3

Divide

2^{2 x}

$2^{2 x}$ by

1

$1$ .

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

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

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

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

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

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

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

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

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})

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

Step 1.4

Expand

\ln (3^{2 y})

$\ln (3^{2 y})$ by moving

2 y

$2 y$ outside the logarithm.

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

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

Step 1.5

Divide each term in

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

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

2 \ln (3)

$2 \ln (3)$ and simplify.

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Step 1.5.1

Divide each term in

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

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

2 \ln (3)

$2 \ln (3)$ .

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

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Step 1.5.2.1

Cancel the common factor of

2

$2$ .

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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)}

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

Step 1.5.2.1.2

Rewrite the expression.

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

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

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

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

Step 1.5.2.2

Cancel the common factor of

\ln (3)

$\ln (3)$ .

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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)}

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

Step 1.5.2.2.2

Divide

y

$y$ by

1

$1$ .

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