Question

Let log 10! = a. Find an expression for log 9! in terms of a.

Answer 1

Answer:

$log 9! = a-1$

Step-by-step explanation:

$(log10! = a) = (10^{a} = 10!)\\\\\\10! = 10 * 9 * 8 * 7 ...\\\\10^a = 10 * 9 * 8 * 7 ...\\\\10^a = 10 * 9!\\\\\frac{10^a}{10^1} = 9!\\\\10^{a-1} = 9!\\\\log9! = a-1$

Answer 2

The expression for log 9! in terms of a is $log9! = a - 1$.

Given that,

• Let log 10! = a.

Based on the above information, the calculation is as follows:

$(log10! = a) = (10^{a} = 10!)\\\\10! = 10\times 9\times 8\times 7\\\\10^{a} = = 10\times 9\times 8\times 7\\\\10^{a} = 10\times 9!\\\\\frac{10^{a}}{10^{1}} = 9!\\\\10^{a - 1} = 9!\\\\log9! = a - 1$

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