Question

Consider the following equation, ${{2}^{n-7}}\times {{5}^{n-4}}=1250$, then find the value of $n$.

Answer 1

In order to solve this question we have to first calculate the prime factorization of 1250 and then arrange it in the powers of 2 and 5. After arranging, compare the powers of 2 and 5 both sides and get two equations. Get the value of $n$ after solving those equations.

Complete step-by-step solution:
It is given that ${{2}^{n-7}}\times {{5}^{n-4}}=1250$ , now we have to calculate the prime factorization of 1250.
"Prime Factorization" is finding which prime numbers multiply together to make the original number.
Now we have the number 1250.
So, $1250=2\times 5\times 5\times 5\times 5$
It can also be written as, $1250={{2}^{1}}\times {{5}^{4}}$.
Putting the value of 1250 in the form of prime factorization, we get
$\therefore {{2}^{n-7}}\times {{5}^{n-4}}={{2}^{1}}\times {{5}^{4}}$
Comparing the powers of 2 and 5 both sides, we get two equations
$\Rightarrow n-7=1\And n-4=4$
Calculating the values of $n$ from both equations, we get
$\Rightarrow n-7=1\Rightarrow n=7+1=8$
From first equation we get the value of $n=8$.
$\Rightarrow n-4=4\Rightarrow n=4+4=8$
Also, from the second we get the value of $n=8$.
Hence, as we are getting an equal value of $n$ from both the equation, which is equal to 8.
So, $n=8$ is the answer to the given question.

Note: In this type of question, the crux lies in representing the number in RHS in powers of prime numbers, which can be done easily by representing the prime factorization of the number in RHS. After representing in this way, students get 2 or more equations depending on the number of terms in LHS but they only solve one equation and claim that it is the answer. But it is always wise to solve all those equations and get a value of $n$ and if all values are the same then only we can say that this is the correct value, just the way we did in the given question.