Solve for a 4^(a^2)=16^(2a)*64^(2-a)

$4^{a^{2}} = 16^{2 a} \cdot 64^{2 - a}$

Step 1

Rewrite $16$ as $2^{4}$ .

$4^{a^{2}} = {(2^{4})}^{2 a} \cdot 64^{2 - a}$

Step 2

Multiply the exponents in ${(2^{4})}^{2 a}$ .

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

$4^{a^{2}} = 2^{4 (2 a)} \cdot 64^{2 - a}$

Multiply $2$ by $4$ .

$4^{a^{2}} = 2^{8 a} \cdot 64^{2 - a}$

Step 3

Rewrite $64$ as $2^{6}$ .

$4^{a^{2}} = 2^{8 a} \cdot {(2^{6})}^{2 - a}$

Step 4

Multiply the exponents in ${(2^{6})}^{2 - a}$ .

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

$4^{a^{2}} = 2^{8 a} \cdot 2^{6 (2 - a)}$

Apply the distributive property.

$4^{a^{2}} = 2^{8 a} \cdot 2^{6 \cdot 2 + 6 (- a)}$

Multiply $6$ by $2$ .

$4^{a^{2}} = 2^{8 a} \cdot 2^{12 + 6 (- a)}$

Multiply $- 1$ by $6$ .

$4^{a^{2}} = 2^{8 a} \cdot 2^{12 - 6 a}$

Step 5

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

$4^{a^{2}} = 2^{8 a + 12 - 6 a}$

Step 6

Subtract $6 a$ from $8 a$ .

$4^{a^{2}} = 2^{2 a + 12}$

Step 7

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

$2^{2 a^{2}} = 2^{2 a + 12}$

Step 8

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

$2 a^{2} = 2 a + 12$

Step 9

Solve for $a$ .

Subtract $2 a$ from both sides of the equation.

$2 a^{2} - 2 a = 12$

Subtract $12$ from both sides of the equation.

$2 a^{2} - 2 a - 12 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 a^{2} - 2 a - 12$ .

Factor $2$ out of $2 a^{2}$ .

$2 (a^{2}) - 2 a - 12 = 0$

Factor $2$ out of $- 2 a$ .

$2 (a^{2}) + 2 (- a) - 12 = 0$

Factor $2$ out of $- 12$ .

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

Factor $2$ out of $2 (a^{2}) + 2 (- a)$ .

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

Factor $2$ out of $2 (a^{2} - a) + 2 (- 6)$ .

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

Factor.

Factor $a^{2} - a - 6$ 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 $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

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

Remove unnecessary parentheses.

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

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

$a - 3 = 0$

$a + 2 = 0$

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

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

$a - 3 = 0$

Add $3$ to both sides of the equation.

$a = 3$

Set $a + 2$ equal to $0$ and solve for $a$ .

Set $a + 2$ equal to $0$ .

$a + 2 = 0$

Subtract $2$ from both sides of the equation.

$a = - 2$

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

$a = 3, - 2$

Solve for a 4^(a^2)=16^(2a)*64^(2-a)

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