# Logarithimiccs

• Jun 5th 2006, 05:44 AM
scfan000
Logarithimiccs
2 Problems I need help on

Log(5)2/Log(5)X=X/LogX

and

Log4Log(2)X=1

both solved without a calculator
• Jun 5th 2006, 05:49 AM
Jameson
I'm confused about your notation. Is this what you mean for the first one? $\frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x}$?
• Jun 5th 2006, 05:51 AM
scfan000
yes it is
• Jun 5th 2006, 05:58 AM
CaptainBlack
Quote:

Originally Posted by scfan000
2 Problems I need help on

Log(5)2/Log(5)X=X/LogX

and

Log4Log(2)X=1

both solved without a calculator

I think you will have to adopt a clearer notation here.

I'm guessing here, do you intend that

$
Log(5)2=\log_5(2)
$
?

Then do you intend that $LogX$ denote the common (log to base 10)
or natrural log of $X$?

I am also mystified by what you want "Log4Log(2)X" to represent.

Might I suggest that you use "log_n(x)" to denote the log to the base
n of x, be explicit in all instances about the base, and rewrite your questions
accordingly?

RonL
• Jun 5th 2006, 06:13 AM
CaptainBlack
Quote:

Originally Posted by Jameson
I'm confused about your notation. Is this what you mean for the first one? $\frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x}$?

To solve:

$\frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x}$,

first turn it upside down (this is a personal preference, it can be done this way up):

$\frac{\log_{5}(x)}{\log_{5}(2)}=\frac{\log(x)}{x}$,

then apply the change of base rule for logs to the left hand side to reduce
the all the logs to the same base as that on the right hand side.

RonL
• Jun 5th 2006, 06:19 AM
scfan000
i c it now. and i can figure out the problem from there
sry about the notation. i just don't quite know how to yet :)
• Jun 5th 2006, 06:57 AM
naty
haj
You need to solve:
Log(5)2/Log(5)X=X/LogX