Basic Math Examples

\frac{2^{n} + 2^{- n}}{2} = \frac{1 + 4^{n}}{2^{n} + 1}

$\frac{2^{n} + 2^{- n}}{2} = \frac{1 + 4^{n}}{2^{n} + 1}$

Step 1

Take the log of both sides of the equation.

ln (\frac{2^{n} + 2^{- n}}{2}) = ln (\frac{1 + 4^{n}}{2^{n} + 1})

$\ln (\frac{2^{n} + 2^{- n}}{2}) = \ln (\frac{1 + 4^{n}}{2^{n} + 1})$

Step 2

Rewrite

ln (\frac{2^{n} + 2^{- n}}{2})

$\ln (\frac{2^{n} + 2^{- n}}{2})$ as

ln (2^{n} + 2^{- n}) - ln (2)

$\ln (2^{n} + 2^{- n}) - \ln (2)$ .

ln (2^{n} + 2^{- n}) - ln (2) = ln (\frac{1 + 4^{n}}{2^{n} + 1})

$\ln (2^{n} + 2^{- n}) - \ln (2) = \ln (\frac{1 + 4^{n}}{2^{n} + 1})$

Step 3

Rewrite

$\ln (\frac{1 + 4^{n}}{2^{n} + 1})$ as

$\ln (1 + 4^{n}) - \ln (2^{n} + 1)$ .

$\ln (2^{n} + 2^{- n}) - \ln (2) = \ln (1 + 4^{n}) - \ln (2^{n} + 1)$

Step 4

Solve the equation for

$n$ .

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Step 4.1

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{2^{n} + 2^{- n}}{2}) = \ln (1 + 4^{n}) - \ln (2^{n} + 1)$

Step 4.2

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{2^{n} + 2^{- n}}{2}) = \ln (\frac{1 + 4^{n}}{2^{n} + 1})$

Step 4.3

Move all terms containing

$n$ to the left side of the equation.

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Step 4.3.1

Subtract

$\ln (\frac{1 + 4^{n}}{2^{n} + 1})$ from both sides of the equation.

$\ln (\frac{2^{n} + 2^{- n}}{2}) - \ln (\frac{1 + 4^{n}}{2^{n} + 1}) = 0$

Step 4.3.2

Use the quotient property of logarithms,

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{\frac{2^{n} + 2^{- n}}{2}}{\frac{1 + 4^{n}}{2^{n} + 1}}) = 0$

Step 4.3.3

Multiply the numerator by the reciprocal of the denominator.

$\ln (\frac{2^{n} + 2^{- n}}{2} \cdot \frac{2^{n} + 1}{1 + 4^{n}}) = 0$

Step 4.3.4

Multiply

$\frac{2^{n} + 2^{- n}}{2}$ by

$\frac{2^{n} + 1}{1 + 4^{n}}$ .

$\ln (\frac{(2^{n} + 2^{- n}) (2^{n} + 1)}{2 (1 + 4^{n})}) = 0$

Step 4.4

Rewrite

$\ln (\frac{(2^{n} + 2^{- n}) (2^{n} + 1)}{2 (1 + 4^{n})}) = 0$ 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

$b^{y} = x$ .

$e^{0} = \frac{(2^{n} + 2^{- n}) (2^{n} + 1)}{2 (1 + 4^{n})}$

Step 4.5

Cross multiply to remove the fraction.

$(2^{n} + 2^{- n}) (2^{n} + 1) = e^{0} (2 (1 + 4^{n}))$

Step 4.6

Simplify

$e^{0} (2 (1 + 4^{n}))$ .

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Step 4.6.1

Simplify the expression.

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Step 4.6.1.1

Anything raised to

$0$ is

$1$ .

$(2^{n} + 2^{- n}) (2^{n} + 1) = 1 (2 (1 + 4^{n}))$

Step 4.6.1.2

Multiply

$2 (1 + 4^{n})$ by

$1$ .

$(2^{n} + 2^{- n}) (2^{n} + 1) = 2 (1 + 4^{n})$

Step 4.6.2

Apply the distributive property.

$(2^{n} + 2^{- n}) (2^{n} + 1) = 2 \cdot 1 + 2 \cdot 4^{n}$

Step 4.6.3

Multiply

$2$ by

$1$ .

$(2^{n} + 2^{- n}) (2^{n} + 1) = 2 + 2 \cdot 4^{n}$

Step 4.6.4

Multiply

$2 \cdot 4^{n}$ .

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Step 4.6.4.1

Rewrite

$4$ as

$2^{2}$ .

$(2^{n} + 2^{- n}) (2^{n} + 1) = 2 + 2 \cdot {(2^{2})}^{n}$

Step 4.6.4.2

Apply the power rule and multiply exponents,

${(a^{m})}^{n} = a^{m n}$ .

$(2^{n} + 2^{- n}) (2^{n} + 1) = 2 + 2 \cdot 2^{2 n}$

Step 4.6.4.3

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(2^{n} + 2^{- n}) (2^{n} + 1) = 2 + 2^{1 + 2 n}$

Step 4.7

Move all terms containing

$n$ to the left side of the equation.

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Step 4.7.1

Subtract

$2^{1 + 2 n}$ from both sides of the equation.

$(2^{n} + 2^{- n}) (2^{n} + 1) - 2^{1 + 2 n} = 2$

Step 4.7.2

Simplify each term.

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Step 4.7.2.1

Expand

$(2^{n} + 2^{- n}) (2^{n} + 1)$ using the FOIL Method.

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Step 4.7.2.1.1

Apply the distributive property.

$2^{n} (2^{n} + 1) + 2^{- n} (2^{n} + 1) - 2^{1 + 2 n} = 2$

Step 4.7.2.1.2

Apply the distributive property.

$2^{n} \cdot 2^{n} + 2^{n} \cdot 1 + 2^{- n} (2^{n} + 1) - 2^{1 + 2 n} = 2$

Step 4.7.2.1.3

Apply the distributive property.

$2^{n} \cdot 2^{n} + 2^{n} \cdot 1 + 2^{- n} \cdot 2^{n} + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2

Simplify each term.

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Step 4.7.2.2.1

Multiply

$2^{n}$ by

$2^{n}$ by adding the exponents.

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Step 4.7.2.2.1.1

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{n + n} + 2^{n} \cdot 1 + 2^{- n} \cdot 2^{n} + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2.1.2

Add

$n$ and

$n$ .

$2^{2 n} + 2^{n} \cdot 1 + 2^{- n} \cdot 2^{n} + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2.2

Multiply

$2^{n}$ by

$1$ .

$2^{2 n} + 2^{n} + 2^{- n} \cdot 2^{n} + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2.3

Multiply

$2^{- n}$ by

$2^{n}$ by adding the exponents.

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Step 4.7.2.2.3.1

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{2 n} + 2^{n} + 2^{- n + n} + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2.3.2

Add

$- n$ and

$n$ .

$2^{2 n} + 2^{n} + 2^{0} + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2.4

Simplify

$2^{0}$ .

$2^{2 n} + 2^{n} + 1 + 2^{- n} \cdot 1 - 2^{1 + 2 n} = 2$

Step 4.7.2.2.5

Multiply

$2^{- n}$ by

$1$ .

$2^{2 n} + 2^{n} + 1 + 2^{- n} - 2^{1 + 2 n} = 2$

Step 4.8

Move all terms not containing

$n$ to the right side of the equation.

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Step 4.8.1

Subtract

$1$ from both sides of the equation.

$2^{2 n} + 2^{n} + 2^{- n} - 2^{1 + 2 n} = 2 - 1$

Step 4.8.2

Subtract

$1$ from

$2$ .

$2^{2 n} + 2^{n} + 2^{- n} - 2^{1 + 2 n} = 1$

Step 4.9

Rewrite

$2^{1 + 2 n}$ as

$2^{1} \cdot 2^{2 n}$ .

$2^{2 n} + 2^{n} + 2^{- n} - (2 \cdot 2^{2 n}) = 1$

Step 4.10

Rewrite

$2^{2 n}$ as exponentiation.

${(2^{n})}^{2} + 2^{n} + 2^{- n} - (2 \cdot 2^{2 n}) = 1$

Step 4.11

Rewrite

$2^{- n}$ as exponentiation.

${(2^{n})}^{2} + 2^{n} + {(2^{n})}^{- 1} - (2 \cdot 2^{2 n}) = 1$

Step 4.12

Rewrite

$2^{2 n}$ as exponentiation.

${(2^{n})}^{2} + 2^{n} + {(2^{n})}^{- 1} - (2 \cdot {(2^{n})}^{2}) = 1$

Step 4.13

Remove parentheses.

${(2^{n})}^{2} + 2^{n} + {(2^{n})}^{- 1} - 2 {(2^{n})}^{2} = 1$

Step 4.14

Substitute

$u$ for

$2^{n}$ .

$u^{2} + u + u^{- 1} - 2 u^{2} = 1$

Step 4.15

Simplify each term.

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Step 4.15.1

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$u^{2} + u + \frac{1}{u} - 2 u^{2} = 1$

Step 4.15.2

Evaluate the exponent.

$u^{2} + u + \frac{1}{u} - 1 \cdot (2 u^{2}) = 1$

Step 4.15.3

Multiply

$- 1$ by

$2$ .

$u^{2} + u + \frac{1}{u} - 2 u^{2} = 1$

Step 4.16

Subtract

$2 u^{2}$ from

$u^{2}$ .

$- u^{2} + u + \frac{1}{u} = 1$

Step 4.17

Solve for

$u$ .

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Step 4.17.1

Find the LCD of the terms in the equation.

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Step 4.17.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, 1, u, 1$

Step 4.17.1.2

The LCM of one and any expression is the expression.

$u$

Step 4.17.2

Multiply each term in

$- u^{2} + u + \frac{1}{u} = 1$ by

$u$ to eliminate the fractions.

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Step 4.17.2.1

Multiply each term in

$- u^{2} + u + \frac{1}{u} = 1$ by

$u$ .

$- u^{2} u + u \cdot u + \frac{1}{u} u = 1 u$

Step 4.17.2.2

Simplify the left side.

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Step 4.17.2.2.1

Simplify each term.

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Step 4.17.2.2.1.1

Multiply

$u^{2}$ by

$u$ by adding the exponents.

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Step 4.17.2.2.1.1.1

Move

$u$ .

$- (u \cdot u^{2}) + u \cdot u + \frac{1}{u} u = 1 u$

Step 4.17.2.2.1.1.2

Multiply

$u$ by

$u^{2}$ .

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Step 4.17.2.2.1.1.2.1

Raise

$u$ to the power of

$1$ .

$- (u^{1} u^{2}) + u \cdot u + \frac{1}{u} u = 1 u$

Step 4.17.2.2.1.1.2.2

Use the power rule

$a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- u^{1 + 2} + u \cdot u + \frac{1}{u} u = 1 u$

Step 4.17.2.2.1.1.3

Add

$1$ and

$2$ .

$- u^{3} + u \cdot u + \frac{1}{u} u = 1 u$

Step 4.17.2.2.1.2

Multiply

$u$ by

$u$ .

$- u^{3} + u^{2} + \frac{1}{u} u = 1 u$

Step 4.17.2.2.1.3

Cancel the common factor of

$u$ .

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Step 4.17.2.2.1.3.1

Cancel the common factor.

$- u^{3} + u^{2} + \frac{1}{u} u = 1 u$

Step 4.17.2.2.1.3.2

Rewrite the expression.

$- u^{3} + u^{2} + 1 = 1 u$

Step 4.17.2.3

Simplify the right side.

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Step 4.17.2.3.1

Multiply

$u$ by

$1$ .

$- u^{3} + u^{2} + 1 = u$

Step 4.17.3

Solve the equation.

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Step 4.17.3.1

Subtract

$u$ from both sides of the equation.

$- u^{3} + u^{2} + 1 - u = 0$

Step 4.17.3.2

Factor the left side of the equation.

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Step 4.17.3.2.1

Reorder terms.

$- u^{3} + u^{2} - u + 1 = 0$

Step 4.17.3.2.2

Factor out the greatest common factor from each group.

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Step 4.17.3.2.2.1

Group the first two terms and the last two terms.

$(- u^{3} + u^{2}) - u + 1 = 0$

Step 4.17.3.2.2.2

Factor out the greatest common factor (GCF) from each group.

$u^{2} (- u + 1) + 1 (- u + 1) = 0$

Step 4.17.3.2.3

Factor the polynomial by factoring out the greatest common factor,

$- u + 1$ .

$(- u + 1) (u^{2} + 1) = 0$

Step 4.17.3.3

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$- u + 1 = 0$

$u^{2} + 1 = 0$

Step 4.17.3.4

Set

$- u + 1$ equal to

$0$ and solve for

$u$ .

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Step 4.17.3.4.1

Set

$- u + 1$ equal to

$0$ .

$- u + 1 = 0$

Step 4.17.3.4.2

Solve

$- u + 1 = 0$ for

$u$ .

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Step 4.17.3.4.2.1

Subtract

$1$ from both sides of the equation.

$- u = - 1$

Step 4.17.3.4.2.2

Divide each term in

$- u = - 1$ by

$- 1$ and simplify.

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Step 4.17.3.4.2.2.1

Divide each term in

$- u = - 1$ by

$- 1$ .

$\frac{- u}{- 1} = \frac{- 1}{- 1}$

Step 4.17.3.4.2.2.2

Simplify the left side.

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Step 4.17.3.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{u}{1} = \frac{- 1}{- 1}$

Step 4.17.3.4.2.2.2.2

Divide

$u$ by

$1$ .

$u = \frac{- 1}{- 1}$

Step 4.17.3.4.2.2.3

Simplify the right side.

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Step 4.17.3.4.2.2.3.1

Divide

$- 1$ by

$- 1$ .

$u = 1$

Step 4.17.3.5

Set

$u^{2} + 1$ equal to

$0$ and solve for

$u$ .

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Step 4.17.3.5.1

Set

$u^{2} + 1$ equal to

$0$ .

$u^{2} + 1 = 0$

Step 4.17.3.5.2

Solve

$u^{2} + 1 = 0$ for

$u$ .

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Step 4.17.3.5.2.1

Subtract

$1$ from both sides of the equation.

$u^{2} = - 1$

Step 4.17.3.5.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$u = \pm \sqrt{- 1}$

Step 4.17.3.5.2.3

Rewrite

$\sqrt{- 1}$ as

$i$ .

$u = \pm i$

Step 4.17.3.5.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.17.3.5.2.4.1

First, use the positive value of the

$\pm$ to find the first solution.

$u = i$

Step 4.17.3.5.2.4.2

Next, use the negative value of the

$\pm$ to find the second solution.

$u = - i$

Step 4.17.3.5.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$u = i, - i$

Step 4.17.3.6

The final solution is all the values that make

$(- u + 1) (u^{2} + 1) = 0$ true.

$u = 1, i, - i$

Step 4.18

Substitute

$1$ for

$u$ in

$u = 2^{n}$ .

$1 = 2^{n}$

Step 4.19

Solve

$1 = 2^{n}$ .

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Step 4.19.1

Rewrite the equation as

$2^{n} = 1$ .

$2^{n} = 1$

Step 4.19.2

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

$\ln (2^{n}) = \ln (1)$

Step 4.19.3

Expand

$\ln (2^{n})$ by moving

$n$ outside the logarithm.

$n \ln (2) = \ln (1)$

Step 4.19.4

Simplify the right side.

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Step 4.19.4.1

The natural logarithm of

$1$ is

$0$ .

$n \ln (2) = 0$

Step 4.19.5

Divide each term in

$n \ln (2) = 0$ by

$\ln (2)$ and simplify.

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Step 4.19.5.1

Divide each term in

$n \ln (2) = 0$ by

$\ln (2)$ .

$\frac{n \ln (2)}{\ln (2)} = \frac{0}{\ln (2)}$

Step 4.19.5.2

Simplify the left side.

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Step 4.19.5.2.1

Cancel the common factor of

$\ln (2)$ .

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Step 4.19.5.2.1.1

Cancel the common factor.

$\frac{n \ln (2)}{\ln (2)} = \frac{0}{\ln (2)}$

Step 4.19.5.2.1.2

Divide

$n$ by

$1$ .

$n = \frac{0}{\ln (2)}$

Step 4.19.5.3

Simplify the right side.

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Step 4.19.5.3.1

Divide

$0$ by

$\ln (2)$ .

$n = 0$

Step 4.20

Substitute

$i$ for

$u$ in

$u = 2^{n}$ .

$i = 2^{n}$

Step 4.21

Solve

$i = 2^{n}$ .

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Step 4.21.1

Rewrite the equation as

$2^{n} = i$ .

$2^{n} = i$

Step 4.21.2

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

$\ln (2^{n}) = \ln (i)$

Step 4.21.3

Expand

$\ln (2^{n})$ by moving

$n$ outside the logarithm.

$n \ln (2) = \ln (i)$

Step 4.21.4

Divide each term in

$n \ln (2) = \ln (i)$ by

$\ln (2)$ and simplify.

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Step 4.21.4.1

Divide each term in

$n \ln (2) = \ln (i)$ by

$\ln (2)$ .

$\frac{n \ln (2)}{\ln (2)} = \frac{\ln (i)}{\ln (2)}$

Step 4.21.4.2

Simplify the left side.

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Step 4.21.4.2.1

Cancel the common factor of

$\ln (2)$ .

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Step 4.21.4.2.1.1

Cancel the common factor.

$\frac{n \ln (2)}{\ln (2)} = \frac{\ln (i)}{\ln (2)}$

Step 4.21.4.2.1.2

Divide

$n$ by

$1$ .

$n = \frac{\ln (i)}{\ln (2)}$

Step 4.22

Substitute

$- i$ for

$u$ in

$u = 2^{n}$ .

$- i = 2^{n}$

Step 4.23

Solve

$- i = 2^{n}$ .

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Step 4.23.1

Rewrite the equation as

$2^{n} = - i$ .

$2^{n} = - i$

Step 4.23.2

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

$\ln (2^{n}) = \ln (- i)$

Step 4.23.3

The equation cannot be solved because

$\ln (- i)$ is undefined.

Undefined

Step 4.23.4

There is no solution for

$2^{n} = - i$

No solution

Step 4.24

List the solutions that makes the equation true.

$n = 0, \frac{\ln (i)}{\ln (2)}$