Demonstrate that:

$\displaystyle \

\frac{1}{{\sum\limits_{k = 1}^n {\left( {\log _{a_k } x} \right)^{ - 1} } }} = \log _{a_1 *...*a_n } x.

\

$

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- Sep 20th 2008, 04:46 AM #1

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- Sep 20th 2008, 10:54 AM #2
argh! it's hard for me to give a substantial hint here, so i will let you off the hook this time and give you a full solution. you can, however, take the high road and read my sketchy hint and try to do the problem yourself before looking at my solution.

hint: start with the RHS of your equation and equate it to a single variable, say, c. solve for c. now it becomes your objective to show that c is equal to the LHS. the change of base formula, formula (1) below, and the inverse log identity (i am not sure if that is the name, but it is formula (2) below) in your algebraic manipulation.

formula (1): $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$ ...........this is the change of base formula

formula (2): $\displaystyle \log_a b = \frac 1{\log_b a}$ ..............a nice identity that i like. i don't see it shown in class often, which is why i don't know the name.

of course the definition of a log (which you should know by heart) comes in handy as well: $\displaystyle \log_a b = c \Longleftrightarrow a^c = b$ ......call this formula (3). it goes without saying that you should know the other basic laws as well, like how the log of a product is just the sum of the logs etc, i will use these rules without warning.

**Solution:**

Let $\displaystyle \log_{a_1 \cdots a_n} x = c$

$\displaystyle \Rightarrow (a_1 \cdots a_n)^c = x$ .............by formula (3)

$\displaystyle \Rightarrow \log_c (a_1 \cdots a_n)^c = \log_c x$ ...........logged both sides

$\displaystyle \Rightarrow c \log_c (a_1 \cdots a_n) = \log_c x$

$\displaystyle \Rightarrow c ( \log_c a_1 + \cdots + \log_c a_n) = \log_c x$

$\displaystyle \Rightarrow c = \frac {\log_c x}{ \log_c a_1 + \cdots + \log_c a_n}$

now we need to change the right side of this equation into the left side of our original equation, and thus prove the equality. again, if you read the solution up to this point, it is not too late to try and take initiative and solve the rest yourself. moving on

$\displaystyle \Rightarrow c = \displaystyle \frac 1{\frac 1{\log_c x} ( \log_c a_1 + \cdots + \log_c a_n)}$

$\displaystyle \displaystyle = \frac 1{\frac {\log_c a_1}{\log_c x} + \cdots + \frac {\log_c a_n}{\log_c x}}$

$\displaystyle \displaystyle = \frac 1{\log_x a_1 + \cdots + \cdots \log_x a_n}$ ..............by formula (1)

$\displaystyle \displaystyle = \frac 1{\frac 1{\log_{a_1} x} + \cdots + \frac 1{\log_{a_n} x}}$ ................by formula (2)

$\displaystyle \displaystyle = \frac 1{\sum \limits_{k = 1}^n \frac 1{\log_{a_k} x}}$

as desired

- Sep 20th 2008, 11:10 AM #3
$\displaystyle \frac{1}{\sum\limits_{k=1}^{n}\frac{1}{\mathrm{log }_{a_{k}}(x)}}=\frac{1}{\sum\limits_{k=1}^{n}\math rm{log}_{x}(a_{k})}$

................$\displaystyle =\frac{1}{\mathrm{log}_{x}\bigg(\prod\limits_{k=1} ^{n}a_{k}\bigg)}$

................$\displaystyle =\mathrm{log}_{\prod\nolimits_{k=1}^{n}a_{k}}(x),$

which is the desired result.

- Sep 20th 2008, 11:17 AM #4

- Sep 20th 2008, 12:09 PM #5

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- Sep 20th 2008, 01:43 PM #6

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Hello, Ortega!

Demonstrate that: .$\displaystyle \frac{1}{{\sum\limits_{k = 1}^n {\left( {\log _{a_k } x} \right)^{ - 1} } }} = \log _{(a_1\cdot a_2\cdots a_n) } x $

A very convenient variation of the Base-change Formula is: .$\displaystyle \log_a(b) \;=\;\frac{1}{\log_b(a)} $

The left side is: .$\displaystyle \frac{1}{\sum\limits^n_{k=1}\frac{1}{\log_{a_k}(x) }} \;=\;\frac{1}{\sum\limits^n_{k=1}\log_x(a_k)}

$

. . $\displaystyle = \;\frac{1}{\log_x(a_1) + \log_x(a_2) + \log_x(a_3) + \cdots \log_x(a_n)} $

. . $\displaystyle =\;\frac{1}{\log_x(a_1\cdot a_2\cdot a_3\cdots a_n)} $

. . $\displaystyle = \;\log_{(a_1\cdot a_2\cdot a_3\cdots a_n)} x $

- Sep 22nd 2008, 01:01 PM #7

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- Sep 22nd 2008, 08:45 PM #8