# finding logs in terms of another log

• Mar 18th 2009, 10:48 AM
jlefholtz
finding logs in terms of another log
I did several of these but am confused by this one.

let a= log2 5 and b= log3 2. Express the following in terms of a, b, or both.

log2 6=? and log5 3=?

cancel the first one, I figured it out. I am now trying (and trying) the log5 3= (and that 5 is supposed to be the base of the log...I just realized that there's html math commands but haven't used them yet.)
• Mar 18th 2009, 11:33 AM
earboth
Quote:

Originally Posted by jlefholtz
I did several of these but am confused by this one.

let a= log2 5 and b= log3 2. Express the following in terms of a, b, or both.

log2 6=? and log5 3=?

1. $\displaystyle b = \log_3(2)~\implies~\dfrac1b = \log_2(3)$

$\displaystyle \log_2(6)=\log_2(2\cdot 3)=\log_2(2)+\log_2(3)=1+\dfrac1b$

2. $\displaystyle \log_5(3)=\log_5\left(\dfrac62\right)=\log_5(6)-\log_5(2) = \dfrac{\log_2(6)}{\log_2(5)}-\dfrac1a=\dfrac{1+\dfrac1b}a-\dfrac1a$
• Mar 18th 2009, 11:38 AM
jlefholtz
thanks!
Oh thanks! I was going to try using division in the second one but didn't know if it was breaking a rule of logarithms. I can do it from here!
• Mar 18th 2009, 11:54 AM
Soroban
Hello, jlefholtz!

These require some Olympic-level gymnastics . . .

Quote:

Let: .$\displaystyle a\:=\:\log_2(5)\,\text{ and }\,b\:=\:\log_3(2)$

Express the following in terms of $\displaystyle a$ and/or $\displaystyle b$.

$\displaystyle (1)\;\log_2(6)$

$\displaystyle (2)\;\log_5(3)$

Theorem: .$\displaystyle \log_b(a) \:=\:\frac{1}{\log_a(b)}$

(1) We have: .$\displaystyle \log_2(6) \:=\:\log_2(3\cdot2) \:=\:\log_2(3) + \log_2(2)$

. . . . Hence: .$\displaystyle \log_2(6) \:=\:\log_2(3) + 1$ .[1]

. . .We are given: .$\displaystyle \log_3(2) \:=\:b \quad\Rightarrow\quad \frac{1}{\log_2(3)} \:=\:b \quad\Rightarrow\quad \log_2(3) \:=\:\frac{1}{b}$

. . .Substitute into [1]: .$\displaystyle \log_2(6) \:=\:\frac{1}{b} + 1 \quad\Rightarrow\quad\boxed{ \log_2(6)\;=\;\frac{1+b}{b}}$

(2) We are given: .$\displaystyle \begin{array}{ccccccc}\log_2(5) \:=\:a & \Longrightarrow & 2^a \:=\: 5 & {\color{blue}[2]}\\ \log_3(2) \:=\: b & \Longrightarrow & 3^b \:=\: 2 & {\color{blue}[3]}\end{array}$

Raise [3] to the power $\displaystyle a\!:\;\;(3^b)^a \:=\:2^a \quad\Rightarrow\quad 3^{ab} \:=\:2^a$

But from [1]: .$\displaystyle 2^a \:=\:5$

. . So we have: .$\displaystyle 3^{ab} \:=\:5$

$\displaystyle \text{Take logs (base 3): }\;\log_3(3^{ab}) \:=\:\log_3(5) \quad\Rightarrow\quad ab\cdot\underbrace{\log_3(3)}_{\text{This is 1}} \:=\:\log_3(5)$

We have: .$\displaystyle \log_3(5) \:=\:ab \quad\Rightarrow\quad \frac{1}{\log_5(3)} \:=\:ab \quad\Rightarrow\quad \boxed{\log_5(3) \:=\:\frac{1}{ab}}$