Conversion of logarithm eqns: Prove lg(x/n)^lg x = lg(x/n)^n same as (lg x - n)...

[RolandHP]

The end goal here is to prove

is the same as

First I solved

for x and ended up with

(apologies for not providing my full approach to this, I'm a bit short in time)

Then I converted

to

I wrote down

as

and here is where I hit a wall.

need to become

.
I might overlook something ovbious, but I can't come past this part.

I hope I wrote down everything correctly and provided enough information, thanks for all future replies.

 

[stapel]

The end goal here is to prove

is the same as

I'm not clear on what you mean by the first equation above. Do you intend either of the following?

. . . . .\(\displaystyle \mbox{a) }\, \log^{\log(x)}\left(\dfrac{x}{n}\right)\, =\, \log^n\left(\dfrac{x}{n}\right)\)

. . . . .\(\displaystyle \mbox{b) }\, \log \left( \left( \dfrac{x}{n} \right)^{\log(x)}\right)\, =\, \log\left(\left(\dfrac{x}{n}\right)^n\right)\)

First I solved

for x and ended up with

(apologies for not providing my full approach to this, I'm a bit short in time)

Because you haven't shown your steps, I can't figure out which, if either, of the equations is correct, nor can I comment on your result. Sorry.

 

[mmm4444bot]

The end goal here is to prove

is the same as

First I solved

for x …

This tells me that you are not using function notation and that choice (b) in stapel's post is the correct interpretation.

… ended up with

… Then I converted

to

I wrote down

…

You're back where you started; this approach seems circular.


Here's a different way to begin; I am using function notation:

lg[(x/n)^lg(x)] = lg[(x/n)^n]

Subtract, to get zero:

lg[(x/n)^lg(x)] - lg[(x/n)^n] = 0

Apply a property of logarithms:

lg[(x/n)^lg(x) / (x/n)^n] = 0

Next, apply a property of exponents, to simplify the expression in green.

Two more applications of logarithm properties after that, and you're done. :cool:

 

[RolandHP]

This tells me that you are not using function notation and that choice (b) in stapel's post is the correct interpretation.


You're back where you started; this approach seems circular.


Here's a different way to begin; I am using function notation:

lg[(x/n)^lg(x)] = lg[(x/n)^n]

Subtract, to get zero:

lg[(x/n)^lg(x)] - lg[(x/n)^n] = 0


Apply a property of logarithms:

lg[(x/n)^lg(x) / (x/n)^n] = 0

Next, apply a property of exponents, to simplify the expression in green.

Two more applications of logarithm properties after that, and you're done. :cool:

Thank you both for your anwsers, and apologies for not writing down clear enough instructions.