Find the value of $X$ which satisfies the equation $5^{2 x}-6 \times 5^x+5=0$
Answer (2)
Substitute y = 5 x to transform the equation into a quadratic equation: y 2 − 6 y + 5 = 0 .
Factor the quadratic equation: ( y − 5 ) ( y − 1 ) = 0 , yielding y = 5 or y = 1 .
Solve for x using the substitution: 5 x = 5 gives x = 1 , and 5 x = 1 gives x = 0 .
The solutions are x = 0 and x = 1 , so the values of X are 0 , 1 .
Explanation
Recognizing the Quadratic Form We are given the equation 5 2 x − 6 × 5 x + 5 = 0 and we need to find the value of x that satisfies it. This looks like a quadratic equation in disguise! Let's make a substitution to see it more clearly.
Substitution Let y = 5 x . Then, the equation becomes y 2 − 6 y + 5 = 0 . Now it's a standard quadratic equation that we can solve by factoring.
Factoring the Quadratic We can factor the quadratic equation as ( y − 5 ) ( y − 1 ) = 0 . This gives us two possible values for y : y = 5 or y = 1 .
Finding x Now we need to find the values of x that correspond to these values of y . Remember that y = 5 x . So, if y = 5 , we have 5 x = 5 , which means x = 1 . If y = 1 , we have 5 x = 1 , which means x = 0 (since any number raised to the power of 0 is 1).
Solutions Therefore, the values of x that satisfy the equation are x = 0 and x = 1 .
Final Answer The problem asks for the value of X , implying there is only one solution. However, we found two solutions: x = 0 and x = 1 . Since the problem does not specify any further conditions, we can assume either solution is acceptable. Let's state both solutions explicitly. The values of x that satisfy the equation are 0 and 1 .
Examples
Imagine you're tracking the growth of a bacteria population. If the population size at time 'x' is given by 5 x , and you want to find the times when the population reaches certain key milestones (like 1 or 5), you'd solve an equation similar to the one above. This type of exponential equation is fundamental in modeling growth and decay processes in biology, finance, and other fields. Understanding how to solve these equations allows you to predict and control these processes effectively.
To solve the equation 5 2 x − 6 × 5 x + 5 = 0 , we substitute y = 5 x to transform it into a quadratic equation, leading to the factors ( y − 5 ) ( y − 1 ) = 0 . This gives us two values for y , which correspond to x = 1 and x = 0 . Therefore, the solutions are x = 0 and x = 1 .
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