Share
Solving for a in an equation with factorials and exponents
Published on October 23, 2024
Algebra
The solution to the equation 3^(a^2 - 10a + 25)! = 729^4 is a = 3 or a = 7, found by equating exponents after expressing both sides with the same base and solving the resulting quadratic equation.
Question
If \[ 3^{\left(a^{2}-10 a+25\right)!}=729^{4} \] Then, \[ a=\text { ? } \]
Answer
Let’s solve this equation step-by-step.
1. Express both sides of the equation with the same base.
Notice that 729 is a power of 3: 729 = 3<sup>6</sup>. Substitute this into the equation:
Simplify the right side using the rule (x<sup>m</sup>)<sup>n</sup> = x<sup>m*n</sup>:
2. Equate the exponents.
Since the bases are the same, we can equate the exponents:
3. Find the factorial.
Recall that 24 is the factorial of 4 (4! = 432*1 = 24). Therefore:
4. Solve for a.
For the factorials to be equal, the expressions inside the factorials must be equal:
5. Solve the quadratic equation.
Subtract 4 from both sides:
Factor the quadratic equation:
This gives us two possible solutions:
Therefore, the possible values of a are 3 and 7.