# Thread: Proving Logs

1. ## Proving Logs

Prove:
[attached] ...
I have no idea on how to even approach this question, since i only know how to prove ones with the same base.
thanks.

2. Originally Posted by incognito_301
Prove:
[attached] ...
I have no idea on how to even approach this question, since i only know how to prove ones with the same base.
thanks.

Hint: Here are the formulas you need to know:

(1) The change of base formula: $\displaystyle \log_a b = \frac {\log_c b}{\log_c a}$

(2) Cor. of (1): $\displaystyle \log_a b = \frac 1{\log_b a}$ ....(by taking $\displaystyle c = b$ in part (1))

$\displaystyle a$ and $\displaystyle b$ are general terms here, not the $\displaystyle a$ and $\displaystyle b$ in your problem.

Apply the above formulas in the order they are given and you will have your proof. just to give you a direction (the manipulation can work in either) start with the left hand side of your equation

3. ok so I got this:
[attached]
now what?
cant think of a single thing to do now

4. Originally Posted by incognito_301
ok so I got this:
[attached]
now what?
cant think of a single thing to do now
you are not listening to what i am saying. i said apply the rules in the order i gave you. clearly you applied the second one first.

look at where you are starting and where you want to get. x and y are not the base of any log here, why did you make them the bases? you always want the bases to be a or b. now you want to start with an expression that has $\displaystyle \log_a x$ and $\displaystyle \log_b y$ and end up with an expression that has $\displaystyle \log_a y$ and $\displaystyle \log_b x$

try again

5. Hello, incognito_301!

I must assume that you know the Base-Change Formula:

. . . $\displaystyle \log_Q(P) \:=\:\frac{\log(P)}{\log(Q)}$

Prove: .$\displaystyle \log_a(x)\cdot\log_b(y) \:=\:\log_a(y)\cdot\log_b(x)$

The left side is:

. . $\displaystyle \log_a(x)\cdot\log_b(y) \:=\:\frac{\log(x)}{\log(a)}\cdot\frac{\log(y)}{\l og(b)} \;=\;\frac{\log(y)}{\log(a)}\cdot\frac{\log(x)}{\l og(b)}$ .$\displaystyle = \;\log_a(y)\cdot\log_b(x)$

6. wow, this was so easy, i'm almost ashamed that i asked it in the first place. lol
this is truly the best help-forum that i've come across: if you do your best to solve a question, people will help you. amazing forum. thank you.

7. Originally Posted by Soroban
Hello, incognito_301!

I must assume that you know the Base-Change Formula:

. . . $\displaystyle \log_Q(P) \:=\:\frac{\log(P)}{\log(Q)}$

The left side is:

. . $\displaystyle \log_a(x)\cdot\log_b(y) \:=\:\frac{\log(x)}{\log(a)}\cdot\frac{\log(y)}{\l og(b)} \;=\;\frac{\log(y)}{\log(a)}\cdot\frac{\log(x)}{\l og(b)}$ .$\displaystyle = \;\log_a(y)\cdot\log_b(x)$

Nice solution.

This is the one i was thinking of

$\displaystyle \log_a x \cdot \log_b y = \frac {\log_b x}{\log_b a} \cdot \frac {\log_a y}{\log_a b}$ ..........by formula (1) in post #2

...................$\displaystyle = \log_b x \cdot \log_a y$

(since $\displaystyle \log_b a \cdot \log_a b = 1$ by formula (2) in post #2)