Better to provide a little explanation for the stuff(even if it’s completely wrong) so, that we can exactly pinpoint at which step you’re making the mistake.

The theory part which you need to know to solve this is:

In our decimal number system, a number terminates if its denominator (when written in simplest form) has only 2 and/or 5 as prime factors. This is because 10, which is the base of our number system, is the product of 2 and 5.

For example, the fraction 1/2 = 0.5 terminates because the denominator (2) is a factor of 10. Similarly, 3/5 = 0.6 also terminates because the denominator (5) is a factor of 10.

However, if we have a fraction like 1/3 = 0.333…, it does not terminate because the denominator (3) is not a factor of 10.

So, when we say that for a quotient to terminate with one decimal digit, the denominator must be a factor of 10, we mean that the denominator should be composed only of the prime factors of 10 (which are 2 and/or 5). If it has any other prime factors, like in the case of 1/3, then it will not terminate.

Now, this is how we solve it:

The expression is \dfrac{3^x * 5^2} { (3^5 * 5^3)} .

This simplifies to \dfrac{3^{(x-5)} }{ 5}.

For the quotient to terminate with one decimal digit, the denominator must be a factor of 10 (since 10 is the base of our number system). The only way for this to happen is if the denominator is either 2, 5, or a power of 10.

Since our denominator is `5`

, it’s already a factor of 10, so we don’t need to worry about the 3^{(x-5)} term affecting the denominator.

Therefore, any value of `x`

that is greater than or equal to `5`

will satisfy the condition, because it ensures that 3^{(x-5)} is an integer and does not contribute any additional factors of `2`

or `5`

to the denominator.