Finding The Logs

Algebra Level 2

$\large \begin{cases} a = \log_x 2 \\ b = \log_x 3 \\ c = \log_x 5 \end{cases}$

What is the value of $\log_x (1\times2\times3\times4\times5\times6)$ in terms of $a$ , $b$ and $c$ as defined above?

4a + 2b + c

12a + 9b + 5c

3a + b + c

2a + 3b + 5c

3 solutions

Chew-Seong Cheong
Aug 7, 2016

Relevant wiki: Properties of Logarithms - Basic

$\begin{aligned} \log_x (6!) & = \log_x (2^4 \cdot 3^2 \cdot 5) \\ & = 4 \log_x 2 + 2 \log_x 3 + \log_x 5 \\ & = \boxed{4a+2b+c} \end{aligned}$

Moderator note:

Yes. The reason why we write $6!$ as $2^4 \times 3^2 \times 5$ is because we want to write $\log_x (1\times2\times3\times4\times5\times6)$ as a linear combination of $\log_x 2, \log_x 3$ and $\log_x 5$ .

Bonus question : What would the answer be if the conditions $\begin{cases} a = \log_x 2 \\ b = \log_x 3 \\ c = \log_x 5 \end{cases}$ were replaced by $\begin{cases} a = \log_2 x \\ b = \log_3 x \\ c = \log_5 x \end{cases}$ ?

A Former Brilliant Member
Aug 13, 2016

Let $a = \log_{x}2$ , $b = \log_{x}3$ and $c = \log_{x}5$ . Then:

$\begin{aligned} \log_{x}(6!) &= \log_{x}(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2) \\ &= \log_{x}6 + \log_{x}5 + \log_{x}4 + \log_{x}3 + \log_{x}2 \\ &= \log_{x}(2 \cdot 3) + \log_{x}5 + \log_{x}(2 \cdot 2) + \log_{x}3 + \log_{x}2 \\ &= \log_{x}2 + \log_{x}3 + \log_{x}5 + \log_{x}2 + \log_{x}2 + \log_{x}3 + \log_{x}2 \\ &= 4\log_{x}2 + 2\log_{x}3 + \log_{x}5 \\ &= 4a + 2b + c \end{aligned}$

as required.

Moderator note:

Great rendition of your work! You have utilized the property $\log a + \log b = \log (a \times b)$ well!

Bonus question : If this time, we are given the numerical value of $x$ , is it always true that $a<b<c$ ? Why or why not?

J D
Aug 7, 2016

Should be 4a + b + c

Moderator note:

Can you explain how you arrived at the answer? Did you remember to account for $6 = 2 \times 3$ ?

Finding The Logs

3 solutions

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