What is the value of $x$ $x$ in $log (x + 2)! - log (x + 1)! = 2$ $log(x+ 2)!- log(x + 1)! = 2$ ?

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Answer 1 · 2016-12-18T18:33:25

$x = 98$ $x = 98$

Use the rule ${log}_{a} n - {log}_{a} m = {log}_{a} (\frac{n}{m})$ $color(magenta)(log_a n - log_a m = log_a (n/m)$ .

$\frac{log ((x + 2)!)}{(x + 1)!} = 2$ $log ((x + 2)!)/((x + 1)!) = 2$

Expand the factorials using the definition $color(magenta)(n! = n(n - 1)(n - 2)(...)(1)$

$((x + 2)(x + 1)!)/((x+ 1)!) = 10^2$

Eliminate the factorials, to be left with:

$x + 2 = 100$

$x = 98$

Hopefully this helps!

What is the value of xxx in log(x+ 2)!- log(x + 1)! = 2log(x+2)!−log(x+1)!=2log(x+ 2)!- log(x + 1)! = 2?