Show that 2/log₉ A - 1/log₃ A = 3/log₃ A

[thomasrules]

I don't know how to show that:

2/log underscore9 A-1/log underscore3 A = 3/log underscore3 A

 

[Jameson]

Are you asking:

$$\log_{9}(a-1) - \log_{3}(a) = 3 \log_{3}(a)$$

?

 

[thomasrules]

NO.

2/((log under9)a)-((1/log under3)a)= 3/((log under3)a)

GOT IT?

 

[vincentchan]

your question is basically
$$\frac{2}{log_9 A} - \frac{1}{log_3A} = \frac{3}{log_3A}$$

do you know the identity:
$$log_ax=\frac{log_bx}{log_ba}$$


this implies
log_9A= (1/2)log_3A [/color]

 

[thomasrules]

yes

and so...

 

[Jameson]

Ok. So you have $$\frac{2}{log_9 A} - \frac{1}{log_3A} = \frac{3}{log_3A}$$

If you get a common numerator of 6, you can rewrite this as

$$\frac{6}{3log_9 A} - \frac{6}{6log_3A} = \frac{6}{2log_3A}$$

From there you can multiply by $$\frac{1}{6}$$ and cancel out the top.

Can you solve it from there?

 

[vincentchan]

see the white letter in my first post

 

[James R]

Probably, you need the fact that

$$2\log_9 x = \log_3 x$$

which follows because:

If we put $$y = \log_9 x$$, then $$x = 9^y$$, and therefore

$$\log_3 x = \log_3 9^y = y\log_3 9 = 2y = 2\log_9 x$$

(I hope that's right.)

 

[thomasrules]

nm holy ****...how would I know to set y to that?

and so how does 2log_9x=log_3x?

 

[thomasrules]

damnit all of you gave me different ways but now I'm confused ...can someone really go easy step by step

I tried something else. DOn't know if its right but how do I prove:

log_9a=2log_3a

 

[vincentchan]

plug in $$log_9A=1/2 log_3A$$ in your left hand side of $$\frac{2}{log_9 A} - \frac{1}{log_3A} = \frac{3}{log_3A}$$
you'll see the answer instantly, what is your problem?

the prove had already provided by James R in #8 post, which part you don't understand?

 

[thomasrules]

how you got 1/2log_3A

 

[vincentchan]

how you got 1/2log_3A

see post number 8 by JamesR
or use the identity:
$$log_ax=\frac{log_bx}{log_ba}$$
this identity can be proved by the same method in post #8