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

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

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 $\text{?} = -2$.