help with proofing logarithms!

[siobhanfowler]

Show that the following statement is true:
3/log_2a - 2/log_4a = 1/log_1/2a

help!
okay, because it's division, i would change the logarithms to subtraction?
log_2(3)-log_2(a)- log_4(2)log_4(a)-log_1/2(1)-log_1/2(a)

now what ?
i'm totally lost

the '_' represent subscript of the following number.

 

[tkhunny]

No, silly. It doesn't work that way. Division INSIDE goes to subtraction OUTSIDE. Division OUTSIDE doesn't do anything.

If \(\displaystyle log_{2}(a) = c\), then \(\displaystyle 2^{c} = a\)

Why do I tell you this?

If \(\displaystyle log_{4}(a) = d\), then \(\displaystyle 4^{d} = 2^{2d} = a\)

Again, why do I want to know this?

If \(\displaystyle log_{1/2}(a) = b\), then \(\displaystyle (1/2)^{b} = 2^{-b} = a\)

Just some hints. I wasn't getting it until I started playing with it. Now I begin to see it.

Let's also observe this:

\(\displaystyle log_{2}(8) = 3\) -- Since 2^3 = 8

\(\displaystyle log_{4}(16) = 2\) -- Since 4^2 = 16

\(\displaystyle log_{1/2}(1/2) = 1\) -- Since, well, you get the picture.

NOW we're getting somewhere!

\(\displaystyle \frac{log_{2}(8)}{log_{2}(a)} - \frac{log_{4}(16)}{log_{4}(a)} = \frac{log_{1/2}(1/2)}{log_{1/2}(a)}\)

Remember your change of base property?

\(\displaystyle log_{a}(8) - log_{a}(16) = log_{a}(1/2)\)

If you wish...

\(\displaystyle 3\cdot log_{a}(2) - 4\cdot log_{a}(2) = -log_{a}(2)\)

You're not going to make me do ALL the work, are you?

 

[soroban]

Hello, siobhanfowler!

\(\displaystyle \text{Prove: }\:\frac{3}{\log_2\!a} - \frac{2}{\log_4\!a} \:=\: \frac{1}{\log_{\frac{1}{2}}\!a}\)

\(\displaystyle \text{We can use this identity: }\;\boxed{\log_b\!a \;\equiv\;\frac{1}{\log_a\!b}}\)


\(\displaystyle \text{The left side is: }\:\frac{3}{\log_2\!a} - \frac{2}{\log_4\!a} \;\;\equiv\;\; 3\log_a\!2 - 2\log_a\!4 \;\;=\;\;\log_a(2^3) - \log_a(4^2)\)

. . . . . . . . . \(\displaystyle = \;\;\log_a\!8 - \log_a\!16 \;\;=\;\;\log_a\!(\tfrac{8}{16}) \;\;=\;\;\log_a\!(\tfrac{1}{2}) \;\;\equiv\;\;\frac{1}{\log_{\frac{1}{2}}\!a}\)