Log, Not Wood

Algebra Level 2

Find $n$ if $\log (225!) - \log (223!) = 1 + \log (n!)$

Note : The log is taken to base 10 and n! is the product of first n natural numbers.

The answer is 7.

2 solutions

Athiyaman Nallathambi
Aug 26, 2015

We have, $\log(\frac{225!}{223!}) = \log(225\times224) = \log(5040 \times 10)$ which can be simplified to, $\log(7! \times 10) = \log(7!) +\log(10) = 1+\log(7!)$ Comparing with $1+\log(n!)$ , we get, $\Rightarrow n = \boxed{7}$

Good job @Athiyaman Nallathambi . Did the same way. Upvoted!

Nelson Mandela - 5 years, 9 months ago

Thanks @Nelson Mandela .The question was easy anyway.

Athiyaman Nallathambi - 5 years, 9 months ago

Yes, It was. Keep it up anyways.

Nelson Mandela - 5 years, 9 months ago

Good solution, but I did it in other way .

Anish Harsha - 5 years, 9 months ago

Its pretty much the same way. $\log(225\times224) = \log(10\times n!) \Rightarrow n! = \frac{50400}{10} \Rightarrow n! =5040\, \text{so} \, n= \boxed{7}$ I like your solution more than mine btw ;)

Athiyaman Nallathambi - 5 years, 9 months ago

Thanks, I feel mine is easier, but yours is much clear.

Anish Harsha - 5 years, 9 months ago

Anish Harsha
Aug 26, 2015

Using the rules of logarithms and observing that $\log (10)$ to base 10 is 1, we have $225 \times 224 = 10\times n!$ and so $n! = 5040$ .
Since $7! = 5040$ , n = 7 .

Log, Not Wood

The answer is 7.

2 solutions

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