Solve for a (1/9)^a*(1/3)^a=3^(16-a)

${(\frac{1}{9})}^{a} \cdot {(\frac{1}{3})}^{a} = 3^{16 - a}$

Take the log of both sides of the equation.

$\ln ({(\frac{1}{9})}^{a} \cdot {(\frac{1}{3})}^{a}) = \ln (3^{16 - a})$

Rewrite $\ln ({(\frac{1}{9})}^{a} \cdot {(\frac{1}{3})}^{a})$ as $\ln ({(\frac{1}{9})}^{a}) + \ln ({(\frac{1}{3})}^{a})$ .

$\ln ({(\frac{1}{9})}^{a}) + \ln ({(\frac{1}{3})}^{a}) = \ln (3^{16 - a})$

Expand $\ln ({(\frac{1}{9})}^{a})$ by moving $a$ outside the logarithm.

$a \ln (\frac{1}{9}) + \ln ({(\frac{1}{3})}^{a}) = \ln (3^{16 - a})$

Expand $\ln ({(\frac{1}{3})}^{a})$ by moving $a$ outside the logarithm.

$a \ln (\frac{1}{9}) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

Rewrite $\ln (\frac{1}{9})$ as $\ln (1) - \ln (9)$ .

$a (\ln (1) - \ln (9)) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

The natural logarithm of $1$ is $0$ .

$a (0 - \ln (9)) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

Rewrite $\ln (9)$ as $\ln (3^{2})$ .

$a (0 - \ln (3^{2})) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

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

$a (0 - (2 \ln (3))) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

Multiply $2$ by $- 1$ .

$a (0 - 2 \ln (3)) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

Subtract $2 \ln (3)$ from $0$ .

$a (- 2 \ln (3)) + a \ln (\frac{1}{3}) = \ln (3^{16 - a})$

Rewrite $\ln (\frac{1}{3})$ as $\ln (1) - \ln (3)$ .

$a (- 2 \ln (3)) + a (\ln (1) - \ln (3)) = \ln (3^{16 - a})$

The natural logarithm of $1$ is $0$ .

$a (- 2 \ln (3)) + a (0 - \ln (3)) = \ln (3^{16 - a})$

Subtract $\ln (3)$ from $0$ .

$a (- 2 \ln (3)) + a (- \ln (3)) = \ln (3^{16 - a})$

Expand $\ln (3^{16 - a})$ by moving $16 - a$ outside the logarithm.

$a (- 2 \ln (3)) + a (- \ln (3)) = (16 - a) \ln (3)$

Solve the equation for $a$ .

Simplify the left side.

Reorder $a$ and $- 1$ .

$- 2 \cdot (a \ln (3)) - 1 \cdot (a \ln (3)) = (16 - a) \ln (3)$

Rewrite $- 1 a$ as $- a$ .

$- 2 a \ln (3) - a \ln (3) = (16 - a) \ln (3)$

Subtract $a \ln (3)$ from $- 2 a \ln (3)$ .

$- 3 a \ln (3) = (16 - a) \ln (3)$

Simplify the right side.

Apply the distributive property.

$- 3 a \ln (3) = 16 \ln (3) - a \ln (3)$

Reorder $16 \ln (3)$ and $- a \ln (3)$ .

$- 3 a \ln (3) = - a \ln (3) + 16 \ln (3)$

Simplify the left side.

Simplify $- 3 a \ln (3)$ .

Simplify $- 3 \ln (3)$ by moving $3$ inside the logarithm.

$a (- \ln (3^{3})) = - a \ln (3) + 16 \ln (3)$

Raise $3$ to the power of $3$ .

$a (- \ln (27)) = - a \ln (3) + 16 \ln (3)$

Reorder factors in $a (- \ln (27))$ .

$- a \ln (27) = - a \ln (3) + 16 \ln (3)$

Simplify the right side.

Simplify each term.

Simplify $16 \ln (3)$ by moving $16$ inside the logarithm.

$- a \ln (27) = - a \ln (3) + \ln (3^{16})$

Raise $3$ to the power of $16$ .

$- a \ln (27) = - a \ln (3) + \ln (43046721)$

Move all the terms containing a logarithm to the left side of the equation.

$- a \ln (27) + a \ln (3) - \ln (43046721) = 0$

Add $\ln (43046721)$ to both sides of the equation.

$- a \ln (27) + a \ln (3) = \ln (43046721)$

Factor $a$ out of $- a \ln (27) + a \ln (3)$ .

Factor $a$ out of $- a \ln (27)$ .

$a (- 1 \ln (27)) + a \ln (3) = \ln (43046721)$

Factor $a$ out of $a \ln (3)$ .

$a (- 1 \ln (27)) + a (\ln (3)) = \ln (43046721)$

Factor $a$ out of $a (- 1 \ln (27)) + a (\ln (3))$ .

$a (- 1 \ln (27) + \ln (3)) = \ln (43046721)$

Rewrite $- 1 \ln (27)$ as $- \ln (27)$ .

$a (- \ln (27) + \ln (3)) = \ln (43046721)$

Divide each term in $a (- \ln (27) + \ln (3)) = \ln (43046721)$ by $- \ln (27) + \ln (3)$ and simplify.

Divide each term in $a (- \ln (27) + \ln (3)) = \ln (43046721)$ by $- \ln (27) + \ln (3)$ .

$\frac{a (- \ln (27) + \ln (3))}{- \ln (27) + \ln (3)} = \frac{\ln (43046721)}{- \ln (27) + \ln (3)}$

Simplify the left side.

Cancel the common factor of $- \ln (27) + \ln (3)$ .

Cancel the common factor.

$\frac{a (- \ln (27) + \ln (3))}{- \ln (27) + \ln (3)} = \frac{\ln (43046721)}{- \ln (27) + \ln (3)}$

Divide $a$ by $1$ .

$a = \frac{\ln (43046721)}{- \ln (27) + \ln (3)}$

Simplify the right side.

Factor $- 1$ out of $- \ln (27)$ .

$a = \frac{\ln (43046721)}{- \ln (27) + \ln (3)}$

Factor $- 1$ out of $\ln (3)$ .

$a = \frac{\ln (43046721)}{- \ln (27) - 1 (- \ln (3))}$

Factor $- 1$ out of $- \ln (27) - 1 (- \ln (3))$ .

$a = \frac{\ln (43046721)}{- (\ln (27) - \ln (3))}$

Rewrite negatives.

Rewrite $- (\ln (27) - \ln (3))$ as $- 1 (\ln (27) - \ln (3))$ .

$a = \frac{\ln (43046721)}{- 1 (\ln (27) - \ln (3))}$

Move the negative in front of the fraction.

$a = - \frac{\ln (43046721)}{\ln (27) - \ln (3)}$

The result can be shown in multiple forms.

Exact Form:

$a = - \frac{\ln (43046721)}{\ln (27) - \ln (3)}$

Decimal Form:

$a = - 8$

Solve for a (1/9)^a*(1/3)^a=3^(16-a)

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