Write as a Set of Linear Factors 49^x+7^(x+1)-60=0

$49^{x} + 7^{x + 1} - 60 = 0$

Step 1

Rewrite $7^{x + 1}$ as $7^{x} \cdot 7^{1}$ .

$49^{x} + 7^{x} \cdot 7 - 60 = 0$

Step 2

Rewrite $49^{x}$ as ${(7^{2})}^{x}$ .

${(7^{2})}^{x} + 7^{x} \cdot 7 - 60 = 0$

Step 3

Rewrite ${(7^{2})}^{x}$ as ${(7^{x})}^{2}$ .

${(7^{x})}^{2} + 7^{x} \cdot 7 - 60 = 0$

Step 4

Let $u = 7^{x}$ . Substitute $u$ for all occurrences of $7^{x}$ .

Evaluate the exponent.

$u^{2} + u \cdot 7 - 60 = 0$

Move $7$ to the left of $u$ .

$u^{2} + 7 u - 60 = 0$

Step 5

Factor $u^{2} + 7 u - 60$ 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 $- 60$ and whose sum is $7$ .

$- 5, 12$

Write the factored form using these integers.

$(u - 5) (u + 12) = 0$

Step 6

Replace all occurrences of $u$ with $7^{x}$ .

$(7^{x} - 5) (7^{x} + 12) = 0$

Write as a Set of Linear Factors 49^x+7^(x+1)-60=0

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