Solve for a 9^(a^2)*3^(5(a))=27

$9^{a^{2}} \cdot 3^{5 (a)} = 27$

Step 1

Rewrite $9$ as $3^{2}$ .

${(3^{2})}^{a^{2}} \cdot 3^{5 (a)} = 27$

Step 2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$3^{2 a^{2}} \cdot 3^{5 (a)} = 27$

Step 3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3^{2 a^{2} + 5 (a)} = 27$

Step 4

Create equivalent expressions in the equation that all have equal bases.

$3^{2 a^{2} + 5 (a)} = 3^{3}$

Step 5

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$2 a^{2} + 5 (a) = 3$

Step 6

Solve for $a$ .

Subtract $3$ from both sides of the equation.

$2 a^{2} + 5 a - 3 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

Factor $5$ out of $5 a$ .

$2 a^{2} + 5 (a) - 3 = 0$

Rewrite $5$ as $- 1$ plus $6$

$2 a^{2} + (- 1 + 6) a - 3 = 0$

Apply the distributive property.

$2 a^{2} - 1 a + 6 a - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 a^{2} - 1 a) + 6 a - 3 = 0$

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

$a (2 a - 1) + 3 (2 a - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 a - 1$ .

$(2 a - 1) (a + 3) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 a - 1 = 0$

$a + 3 = 0$

Set $2 a - 1$ equal to $0$ and solve for $a$ .

Set $2 a - 1$ equal to $0$ .

$2 a - 1 = 0$

Solve $2 a - 1 = 0$ for $a$ .

Add $1$ to both sides of the equation.

$2 a = 1$

Divide each term in $2 a = 1$ by $2$ and simplify.

Divide each term in $2 a = 1$ by $2$ .

$\frac{2 a}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{1}{2}$

Divide $a$ by $1$ .

$a = \frac{1}{2}$

Set $a + 3$ equal to $0$ and solve for $a$ .

Set $a + 3$ equal to $0$ .

$a + 3 = 0$

Subtract $3$ from both sides of the equation.

$a = - 3$

The final solution is all the values that make $(2 a - 1) (a + 3) = 0$ true.

$a = \frac{1}{2}, - 3$

Step 7

The result can be shown in multiple forms.

Exact Form:

$a = \frac{1}{2}, - 3$

Decimal Form:

$a = 0.5, - 3$

Solve for a 9^(a^2)*3^(5(a))=27

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