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In Mathematics / High School | 2025-07-03

$\frac{13 \times 5^b-25 \times 5^{b-2}}{5^{b+2}-5^{b+1}}=\frac{3}{5}$

Asked by at696750

Answer (2)

The equation simplifies to an identity, showing that it holds true for all values of b. Thus, b has no specific solution and can be any real number. The given equation is always valid regardless of the value of b.
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Answered by Anonymous | 2025-07-04

Simplify the numerator: 13 × 5 b − 25 × 5 b − 2 = 12 × 5 b .
Simplify the denominator: 5 b + 2 − 5 b + 1 = 20 × 5 b .
Substitute the simplified expressions into the original equation: 20 × 5 b 12 × 5 b ​ = 5 3 ​ .
Simplify the equation to find that it is an identity, meaning it is true for all values of b : 5 3 ​ = 5 3 ​ . Therefore, there is no specific solution for b .

Explanation

Problem Setup We are given the equation 5 b + 2 − 5 b + 1 13 × 5 b − 25 × 5 b − 2 ​ = 5 3 ​ We want to solve for b .

Simplifying the Equation First, we can rewrite 5 b − 2 as 5 2 5 b ​ = 25 5 b ​ . Then the numerator becomes 13 × 5 b − 25 × 25 5 b ​ = 13 × 5 b − 5 b = 12 × 5 b In the denominator, we can rewrite 5 b + 2 as 5 b × 5 2 = 25 × 5 b and 5 b + 1 as 5 b × 5 = 5 × 5 b . Then the denominator becomes 5 b + 2 − 5 b + 1 = 25 × 5 b − 5 × 5 b = 20 × 5 b So the equation becomes 20 × 5 b 12 × 5 b ​ = 5 3 ​

Solving for b We can simplify the fraction by canceling out 5 b :
20 12 ​ = 5 3 ​ This simplifies to 5 3 ​ = 5 3 ​ Since the equation is always true, it means that the original equation is independent of b . However, we must make sure that the denominator is not zero. Since 5 b + 2 − 5 b + 1 = 20 × 5 b , and 5 b is always positive for any real number b , the denominator is never zero. Therefore, b can be any real number.

Final Answer Since the equation holds true regardless of the value of b , there is no specific solution for b . The equation is an identity, meaning it is true for all values of b .


Examples
This problem demonstrates how algebraic manipulation can simplify complex equations. In real life, similar techniques are used to simplify models in physics, engineering, and economics. For example, when analyzing circuits, simplifying equations can help determine the behavior of the circuit regardless of specific component values, as long as certain relationships hold.

Answered by GinnyAnswer | 2025-07-04