Solve for a g=(1/27)^(a+3)

$g = {(\frac{1}{27})}^{a + 3}$

Step 1

Rewrite the equation as ${(\frac{1}{27})}^{a + 3} = g$ .

${(\frac{1}{27})}^{a + 3} = g$

Step 2

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

$\ln ({(\frac{1}{27})}^{a + 3}) = \ln (g)$

Step 3

Expand the left side.

Expand $\ln ({(\frac{1}{27})}^{a + 3})$ by moving $a + 3$ outside the logarithm.

$(a + 3) \ln (\frac{1}{27}) = \ln (g)$

Rewrite $\ln (\frac{1}{27})$ as $\ln (1) - \ln (27)$ .

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

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

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

Rewrite $\ln (27)$ as $\ln (3^{3})$ .

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

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

$(a + 3) (0 - (3 \ln (3))) = \ln (g)$

Multiply $3$ by $- 1$ .

$(a + 3) (0 - 3 \ln (3)) = \ln (g)$

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

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

Step 4

Simplify the left side.

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

Apply the distributive property.

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

Simplify the expression.

Multiply $- 3$ by $3$ .

$a (- 3 \ln (3)) - 9 \ln (3) = \ln (g)$

Reorder factors in $a (- 3 \ln (3)) - 9 \ln (3)$ .

$- 3 a \ln (3) - 9 \ln (3) = \ln (g)$

Step 5

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

$- 3 a \ln (3) - 9 \ln (3) - \ln (g) = 0$

Step 6

Move all terms not containing $a$ to the right side of the equation.

Add $9 \ln (3)$ to both sides of the equation.

$- 3 a \ln (3) - \ln (g) = 9 \ln (3)$

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

$- 3 a \ln (3) = 9 \ln (3) + \ln (g)$

Step 7

Divide each term in $- 3 a \ln (3) = 9 \ln (3) + \ln (g)$ by $- 3 \ln (3)$ and simplify.

Divide each term in $- 3 a \ln (3) = 9 \ln (3) + \ln (g)$ by $- 3 \ln (3)$ .

$\frac{- 3 a \ln (3)}{- 3 \ln (3)} = \frac{9 \ln (3)}{- 3 \ln (3)} + \frac{\ln (g)}{- 3 \ln (3)}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 a \ln (3)}{- 3 \ln (3)} = \frac{9 \ln (3)}{- 3 \ln (3)} + \frac{\ln (g)}{- 3 \ln (3)}$

Rewrite the expression.

$\frac{a \ln (3)}{\ln (3)} = \frac{9 \ln (3)}{- 3 \ln (3)} + \frac{\ln (g)}{- 3 \ln (3)}$

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

Cancel the common factor.

$\frac{a \ln (3)}{\ln (3)} = \frac{9 \ln (3)}{- 3 \ln (3)} + \frac{\ln (g)}{- 3 \ln (3)}$

Divide $a$ by $1$ .

$a = \frac{9 \ln (3)}{- 3 \ln (3)} + \frac{\ln (g)}{- 3 \ln (3)}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $9$ and $- 3$ .

Factor $3$ out of $9 \ln (3)$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Rewrite the expression.

$a = \frac{3}{- 1} + \frac{\ln (g)}{- 3 \ln (3)}$

Move the negative one from the denominator of $\frac{3}{- 1}$ .

$a = - 1 \cdot 3 + \frac{\ln (g)}{- 3 \ln (3)}$

Multiply $- 1$ by $3$ .

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

Move the negative in front of the fraction.

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

Solve for a g=(1/27)^(a+3)

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