Solve for a 2^(a+1)+2^a=3^(b+2)-3^b

$2^{a + 1} + 2^{a} = 3^{b + 2} - 3^{b}$

Factor out $2^{a}$ from the expression.

$2^{a} (2 + 1)$

Add $2$ and $1$ .

$2^{a} \cdot 3 = 3^{b + 2} - 3^{b}$

Move $3$ to the left of $2^{a}$ .

$3 \cdot 2^{a} = 3^{b + 2} - 3^{b}$

Divide each term in $(3 \cdot 2^{a}) = 3^{b + 2} - 3^{b}$ by $3$ and simplify.

Divide each term in $(3 \cdot 2^{a}) = 3^{b + 2} - 3^{b}$ by $3$ .

$\frac{(3 \cdot 2^{a})}{3} = \frac{3^{b + 2}}{3} + \frac{- 3^{b}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 \cdot 2^{a}}{3} = \frac{3^{b + 2}}{3} + \frac{- 3^{b}}{3}$

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

$2^{a} = \frac{3^{b + 2}}{3} + \frac{- 3^{b}}{3}$

Simplify the right side.

Combine the numerators over the common denominator.

$2^{a} = \frac{3^{b + 2} - 3^{b}}{3}$

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

$\ln (2^{a}) = \ln (\frac{3^{b + 2} - 3^{b}}{3})$

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

$a \ln (2) = \ln (\frac{3^{b + 2} - 3^{b}}{3})$

Rewrite $\ln (\frac{3^{b + 2} - 3^{b}}{3})$ as $\ln (3^{b + 2} - 3^{b}) - \ln (3)$ .

$a \ln (2) = \ln (3^{b + 2} - 3^{b}) - \ln (3)$

Simplify the right side.

Simplify $\ln (3^{b + 2} - 3^{b}) - \ln (3)$ .

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$a \ln (2) = \ln (\frac{3^{b + 2} - 3^{b}}{3})$

Split the fraction $\frac{3^{b + 2} - 3^{b}}{3}$ into two fractions.

$a \ln (2) = \ln (\frac{3^{b + 2}}{3} + \frac{- 3^{b}}{3})$

Simplify each term.

Cancel the common factor of $3^{b + 2}$ and $3$ .

Factor $3$ out of $3^{b + 2}$ .

$a \ln (2) = \ln (\frac{3 \cdot 3^{b + 1}}{3} + \frac{- 3^{b}}{3})$

Cancel the common factors.

Factor $3$ out of $3$ .

$a \ln (2) = \ln (\frac{3 \cdot 3^{b + 1}}{3 (1)} + \frac{- 3^{b}}{3})$

Cancel the common factor.

$a \ln (2) = \ln (\frac{3 \cdot 3^{b + 1}}{3 \cdot 1} + \frac{- 3^{b}}{3})$

Rewrite the expression.

$a \ln (2) = \ln (\frac{3^{b + 1}}{1} + \frac{- 3^{b}}{3})$

Divide $3^{b + 1}$ by $1$ .

$a \ln (2) = \ln (3^{b + 1} + \frac{- 3^{b}}{3})$

Cancel the common factor of $3^{b}$ and $3$ .

Factor $3$ out of $- 3^{b}$ .

$a \ln (2) = \ln (3^{b + 1} + \frac{3 (- 3^{b - 1})}{3})$

Cancel the common factors.

Factor $3$ out of $3$ .

$a \ln (2) = \ln (3^{b + 1} + \frac{3 (- 3^{b - 1})}{3 (1)})$

Cancel the common factor.

$a \ln (2) = \ln (3^{b + 1} + \frac{3 (- 3^{b - 1})}{3 \cdot 1})$

Rewrite the expression.

$a \ln (2) = \ln (3^{b + 1} + \frac{- 3^{b - 1}}{1})$

Divide $- 3^{b - 1}$ by $1$ .

$a \ln (2) = \ln (3^{b + 1} - 3^{b - 1})$

Divide each term in $a \ln (2) = \ln (3^{b + 1} - 3^{b - 1})$ by $\ln (2)$ and simplify.

Divide each term in $a \ln (2) = \ln (3^{b + 1} - 3^{b - 1})$ by $\ln (2)$ .

$\frac{a \ln (2)}{\ln (2)} = \frac{\ln (3^{b + 1} - 3^{b - 1})}{\ln (2)}$

Simplify the left side.

Cancel the common factor of $\ln (2)$ .

Cancel the common factor.

$\frac{a \ln (2)}{\ln (2)} = \frac{\ln (3^{b + 1} - 3^{b - 1})}{\ln (2)}$

Divide $a$ by $1$ .

$a = \frac{\ln (3^{b + 1} - 3^{b - 1})}{\ln (2)}$

Solve for a 2^(a+1)+2^a=3^(b+2)-3^b

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