# simplify logarithm

• Mar 10th 2009, 04:11 PM
NidhiS
simplify logarithm
I was wondering if there's a way to simplify this logarithmic expression

Code:

(3/5)^(log5n)-1
i.e. (3/5) to the power log base 5 of n.
• Mar 10th 2009, 05:51 PM
Jhevon
Quote:

Originally Posted by NidhiS
I was wondering if there's a way to simplify this logarithmic expression

Code:

(3/5)^(log5n)-1
i.e. (3/5) to the power log base 5 of n.

um, what is that -1 that i see? you never mentioned it when you translated your problem into words
• Mar 10th 2009, 06:20 PM
NidhiS
-1's just a separate entity.S so you basically do the first part and subtract one from it
• Mar 10th 2009, 06:44 PM
e^(i*pi)
Quote:

Originally Posted by NidhiS
I was wondering if there's a way to simplify this logarithmic expression

Code:

(3/5)^(log5n)-1
i.e. (3/5) to the power log base 5 of n.

not really:

$\displaystyle (\frac{3}{5})^{log_5{n}} - 1$

If we let this equal a we can rearrange to find n.

$\displaystyle a+1 = 0.6^{log_5{n}}$

$\displaystyle ln{a+1} = log_5{(n)}ln{(0.6)}$

$\displaystyle log_5{(n)} = \frac{ln{(0.6)}}{ln{(a+1)}}$

$\displaystyle n = 5^{\frac{ln{(0.6)}}{ln{(a+1)}}}$

the original expression would be easiest to use
• Mar 10th 2009, 06:45 PM
Soroban
Hello, NidhiS!

I'll take a guess at where that -1 belongs.
Whatever was meant, it doesn't simplify very much.

Using every log-trick I know, I can only rewrite it . . .

Quote:

$\displaystyle \left(\frac{3}{5}\right)^{\log_5\!n-1}$

We have: .$\displaystyle \left(\frac{3}{5}\right)^{\log_5\!n - \log_55} \;=\;\left(\frac{3}{5}\right)^{\log_5(\frac{n}{5}) }$ $\displaystyle = \;\frac{3^{\log_5(\frac{n}{5})}}{5^{\log_5(\frac{n }{5})}} \;=\; \frac{3^{\log_5(\frac{n}{5})}}{\frac{n}{5}}$

I used the Base-change Formula on the exponent:

. . $\displaystyle \frac{3^{\frac{\log_3(\frac{n}{5})}{\log_3\!5}}}{\ frac{n}{5}} \;=\;\frac{\left[3^{\log_3(\frac{n}{5})}\right]^{\frac{1}{\log_3\!5}}}{\frac{n}{5}}$ $\displaystyle = \;\frac{\left(\frac{n}{5}\right)^{\frac{1}{\log_3\ !5}}}{\frac{n}{5}} \;=\;\frac{(\frac{n}{5})^{\log_5\!3}}{\frac{n}{5}} \;=\;\left(\frac{n}{5}\right)^{\log_5\!3-1}$

See? . . . It's no simpler than the original expression.

• Mar 10th 2009, 06:48 PM
Jhevon
Quote:

Originally Posted by NidhiS
I was wondering if there's a way to simplify this logarithmic expression

Code:

(3/5)^(log5n)-1
i.e. (3/5) to the power log base 5 of n.

what was the original problem? why do you want to simplify this. perhaps there is something else you want to get at that we do not have to do this. because it is messy. using the rule $\displaystyle a^{\log_a X} = X$, we can get this down to

$\displaystyle n^{\frac 1{\log_3 5} - 1} - 1$

which is not really simpler, just different. state your intentions and we can take it from there
• Mar 10th 2009, 08:28 PM
NidhiS
$\displaystyle (\frac{3}{5})^{log_5{n}} - 1$

This here is what I want to simplify.I want to simplify it for my algorithm class.(Worried)My professor said that I shouldn't leave it like this nad perhaps should simplify.e^pi*i ,I like your way,but I know there's another way to do it,which has something to do with the property of the common base in the logarithms.
• Mar 10th 2009, 08:58 PM
NidhiS
There's this logarithm property that I know of,which is,

$\displaystyle (a)^{log_b{y}} = (y)^{log_b {a}}$

which i was thinking of using like this,

$\displaystyle (\frac{3}{5})^{log_5{n}}$

changes to

$\displaystyle ({3}.{(5) ^ {-1}})^{log_5{n}} - 1$

Then using the above property I get,

$\displaystyle ({3}.{(5) ^ {-1}})^{log_5{n}} - 1$

$\displaystyle ({3}.{n})^{log_5{-5}} - 1$

which is equal to

$\displaystyle ({3}.{n}) ^ {-1}$

Do you guys this this is fine?Or am I going wrong somewhere?