finding logs in terms of another log

jlefholtz · #1 $Old$ March 18th, 2009, 11:48 AM

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.)

earboth · #2 $Old$ March 18th, 2009, 12:33 PM

Quote:

Originally Posted by jlefholtz $View Post$

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

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

2. $\log_5(3)=\log_5\left(\dfrac62\right)=\log_5(6)-\log_5(2) = \dfrac{\log_2(6)}{\log_2(5)}-\dfrac1a=\dfrac{1+\dfrac1b}a-\dfrac1a$

jlefholtz · #3 $Old$ March 18th, 2009, 12:38 PM

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!

Soroban · #4 $Old$ March 18th, 2009, 12:54 PM

Hello, jlefholtz!

These require some Olympic-level gymnastics . . .

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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