Question

exponential eqns common base (level 3)
score: 1/2 penalty: none
question
solve for all values of x.
\\(10^{3x^2 - 21} = 100^{-12x - 21}\\)
answer attempt 1 out of 2
\\(\oplus\\) additional solution \\(\circleddash\\) no solution
\\(x = \square\\) submit answer

Explanation:

Step1: Rewrite with common base

Note that $100 = 10^2$, so rewrite the right-hand side:
$10^{3x^2 - 21} = (10^2)^{-12x - 21}$

Step2: Simplify exponent on right

Use exponent rule $(a^m)^n = a^{mn}$:
$10^{3x^2 - 21} = 10^{2(-12x - 21)}$
$10^{3x^2 - 21} = 10^{-24x - 42}$

Step3: Set exponents equal

Since bases are equal, exponents are equal:
$3x^2 - 21 = -24x - 42$

Step4: Rearrange to quadratic form

Bring all terms to left side:
$3x^2 + 24x + 21 = 0$
Divide by 3:
$x^2 + 8x + 7 = 0$

Step5: Factor quadratic

Factor the trinomial:
$(x + 1)(x + 7) = 0$

Step6: Solve for x

Set each factor equal to 0:
$x + 1 = 0$ or $x + 7 = 0$

Answer:

$x = -1$, $x = -7$