Solve: 7^(-8) × (1/7)^(5x) = 49 | Exponent Equation Challenge

Question

$7^{-8}\cdot(\frac{1}{7})^{5\cdot\text{?}}=49$

00:00 Fill in the missing exponent

00:03 Let's decompose 42 into 7 squared

00:08 When raising a fraction to a power, raise both the numerator and denominator

00:14 Let's apply this formula to our exercise

00:27 When multiplying powers with equal bases

00:30 The exponent of the result equals the sum of the exponents

00:33 Let's apply this formula to our exercise, we'll then proceed to add together the exponents

00:41 Let's compare the exponents and determine the unknown

00:52 Isolate the unknown

01:09 This is the solution

To solve this problem, we begin by converting each part of the expression into a base that simplifies the comparison.

Step 1: Rewrite $(\frac{1}{7})^{5\cdot ?}$ as $(7^{-1})^{5\cdot ?}$ , which becomes $7^{-5\cdot ?}$ .
Step 2: The left side now reads $7^{-8} \times 7^{-5\cdot ?}$ .
Step 3: Combine the exponents since they have the same base: $7^{-8} \times 7^{-5\cdot ?} = 7^{-8 - 5\cdot ?}$ .
Step 4: The right side of the equation is $49$ , which can be rewritten as $7^2$ .
Step 5: Now we have an equation with the same base on both sides: $7^{-8 - 5\cdot ?} = 7^2$ .
Step 6: Equate the exponents: $-8 - 5\cdot ? = 2$ .
Step 7: Solve for '?':
$-8 - 5\cdot ? = 2$
Add 8 to both sides:
$-5\cdot ? = 2 + 8$
$-5\cdot ? = 10$
Divide by $-5$ :
$? = \frac{10}{-5} = -2$ .

Therefore, the solution to the problem is $\text{?} = -2$ .

$-2$