Solve for x 9^x*3^(x^2)=81^2

$9^{x} \cdot 3^{x^{2}} = 81^{2}$

Step 1

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

${(3^{2})}^{x} \cdot 3^{x^{2}} = 81^{2}$

Step 2

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

$3^{2 x} \cdot 3^{x^{2}} = 81^{2}$

Step 3

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

$3^{2 x + x^{2}} = 81^{2}$

Step 4

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

$3^{2 x + x^{2}} = 3^{4 \cdot 2}$

Step 5

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

$2 x + x^{2} = 4 \cdot 2$

Step 6

Solve for $x$ .

Multiply $4$ by $2$ .

$2 x + x^{2} = 8$

Subtract $8$ from both sides of the equation.

$2 x + x^{2} - 8 = 0$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$2 u + u^{2} - 8 = 0$

Factor $2 u + u^{2} - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

Write the factored form using these integers.

$(u - 2) (u + 4) = 0$

Replace all occurrences of $u$ with $x$ .

$(x - 2) (x + 4) = 0$

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

$x - 2 = 0$

$x + 4 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

Solve for x 9^x*3^(x^2)=81^2

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