Question

resolver ecuaciones reescribiendo un número como una potencia
resolver: 3,125 = 5^(-10 + 3x)

x = \boxed{}

Explanation:

Step1: Convertir 3,125 a potencia de 5

Sabemos que \( 5^3 = 125 \), pero 3,125 es \( \frac{25}{8} \)? No, espera, 3.125 en fracción es \( \frac{25}{8} \)? No, 3.125 = \( \frac{25}{8} \)? No, 3.125 = \( \frac{25}{8} \) es incorrecto. Wait, 5^3 = 125, 5^(-3) = 1/125 = 0.008. Wait, 3.125 = 25/8? No, 3.125 = 25/8? 25 dividido por 8 es 3.125. Pero 5^3 = 125, 5^(-3) = 1/125 = 0.008. Wait, maybe 3.125 is 5^3 / 100? No, no. Wait, 3.125 = 5^3 / 40? No, that's not right. Wait, maybe the number is 3125? Wait, the problem has 3,125 which is 3.125? Or 3125? Wait, 3125 is 5^5, because 5^1=5, 5^2=25, 5^3=125, 5^4=625, 5^5=3125. Ah! Maybe it's a typo, 3,125 (with comma as decimal) or 3125 (without). Let's check: if it's 3125, then 3125 = 5^5. So let's assume it's 3125 (maybe the comma is a thousand separator or decimal, but in Spanish, comma is decimal? Wait, in Spanish, decimal is comma? No, in some countries, decimal is comma. Wait, 3,125 would be 3.125 in English. But 3.125 = 25/8, which is not a power of 5. But 3125 is 5^5. So maybe the problem is 3125 = 5^(-10 + 3x). Let's proceed with 3125 = 5^5.

So \( 3125 = 5^5 \), entonces la ecuación es \( 5^5 = 5^{-10 + 3x} \)

Step2: Igualar exponentes (ya que bases son iguales)

Si \( a^m = a^n \), entonces \( m = n \) (para \( a > 0, a
eq 1 \)). Entonces:

\( 5 = -10 + 3x \)

Step3: Resolver para x

Sumamos 10 a ambos lados:

\( 5 + 10 = 3x \)

\( 15 = 3x \)

Dividimos por 3:

\( x = \frac{15}{3} = 5 \)

Answer:

\( x = 5 \)