2 Problems I need help on

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

and

Log4Log(2)X=1

both solved without a calculator

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- Jun 5th 2006, 05:44 AMscfan000Logarithimiccs
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 AMJameson
I'm confused about your notation. Is this what you mean for the first one? $\displaystyle \frac{\log_{5}2}{\log_{5}x}=\frac{x}{\log x}$?

- Jun 5th 2006, 05:51 AMscfan000
yes it is

- Jun 5th 2006, 05:58 AMCaptainBlackQuote:

Originally Posted by**scfan000**

I'm guessing here, do you intend that

$\displaystyle

Log(5)2=\log_5(2)

$ ?

Then do you intend that $\displaystyle LogX$ denote the common (log to base 10)

or natrural log of $\displaystyle X$?

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 AMCaptainBlackQuote:

Originally Posted by**Jameson**

To solve:

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

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

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

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 AMscfan000
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 AMnatyhaj
You need to solve:

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

In your notation:

Log(5)2 can be written also as Log2/Log5

and similary

Log(5)X = LogX/Log5

When we put this into equation we get:

Log2/LogX=X/LogX

From this: X=Log2