Question

solve the equation.
$7^{x^2 - 35} = 49^x$
the solution set is ( ).
(simplify your answer. use a comma to separate answers as needed.)

Explanation:

Step1: Rewrite base as power of 7

Since \(49 = 7^2\), rewrite \(49^x\) as \((7^2)^x\). Using exponent rule \((a^m)^n = a^{mn}\), we get \((7^2)^x = 7^{2x}\). So the equation \(7^{x^2 - 35} = 49^x\) becomes \(7^{x^2 - 35} = 7^{2x}\).

Step2: Set exponents equal

For exponential functions with the same base (\(a^m = a^n \implies m = n\) when \(a > 0, a
eq 1\)), here \(a = 7\), so set \(x^2 - 35 = 2x\).

Step3: Solve quadratic equation

Rearrange \(x^2 - 2x - 35 = 0\). Factor the quadratic: find two numbers that multiply to \(-35\) and add to \(-2\). These numbers are \(-7\) and \(5\). So \((x - 7)(x + 5) = 0\). Set each factor to zero: \(x - 7 = 0 \implies x = 7\); \(x + 5 = 0 \implies x = -5\).

Answer:

\(7, -5\)