\frac{13 \times 5^b-25 \times 5^{b-2}}{5^{b+2}-5^{b+1}}=\frac{3}{5}
Answer (2)
The given equation simplifies to an identity, confirming that it holds true for all values of b . Therefore, there is no unique solution for b . The solution reflects the consistency and behavior of the equation across all real numbers.
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Simplify the equation by rewriting exponential terms.
Cancel out common factors to obtain a simplified fraction.
Verify that the simplified equation is an identity, true for all values of b .
Conclude that there is no unique solution for b , as the equation holds for all real numbers. no unique solution
Explanation
Problem Setup We are given the equation 5 b + 2 − 5 b + 1 13 × 5 b − 25 × 5 b − 2 = 5 3 and we want to solve for b .
Rewriting Exponents First, let's rewrite the terms with exponents. We can rewrite 5 b − 2 as 5 2 5 b = 25 5 b . Also, 5 b + 2 = 5 b × 5 2 = 25 × 5 b and 5 b + 1 = 5 b × 5 = 5 × 5 b . Substituting these into the equation, we get 25 × 5 b − 5 × 5 b 13 × 5 b − 25 × 25 5 b = 5 3 .
Simplifying the Equation Now, let's simplify the equation. We have 25 × 5 b − 5 × 5 b 13 × 5 b − 5 b = 5 3 . Factoring out 5 b from the numerator and the denominator, we get 5 b ( 25 − 5 ) 5 b ( 13 − 1 ) = 5 3 .
Solving for b We can cancel out 5 b from the numerator and the denominator, which gives us 25 − 5 13 − 1 = 5 3 . Simplifying the fraction, we have 20 12 = 5 3 . Since 20 12 = 5 3 , the equation is true for all values of b . However, we must ensure that the denominator is not zero. Since 5 b + 2 − 5 b + 1 = 5 b + 1 ( 5 − 1 ) = 4 ⋅ 5 b + 1 , the denominator is never zero. Therefore, the equation holds for all real numbers b .
Final Answer Since the equation holds true for all values of b , there is no unique solution for b . However, the question implies that we need to find a specific value for b . The original equation simplifies to 5 3 = 5 3 , which is an identity. This means that the equation is true for any value of b . Therefore, there is no specific value of b that satisfies the equation, as it is true for all b . There is no specific solution for b .
Examples
This problem demonstrates how exponential equations can sometimes simplify to identities, meaning they are true for all values of the variable. In real-world scenarios, this could represent a system where a certain ratio remains constant regardless of the input. For example, in a chemical reaction, if the ratio of reactants to products remains constant over time, the equation describing the reaction would hold true for all time values, similar to how our equation holds true for all values of 'b'. Understanding such identities is crucial in modeling and predicting system behavior in various scientific and engineering applications.
