1. ## Logs

I have two questions:

1. If $log_{14} 7 = a$ and $log_{14} 5 = b$ calculate $log_{35} 28$
2. Let $log 36 = a$ and $log 125 = b$ find $log (1/12)$ in terms of $a$ and $b$

2. Originally Posted by ntrantrinh
I have two questions:

1. If $log_{14} 7 = a$ and $log_{14} 5 = b$ calculate $log_{35} 28$
[snip]
It follows that $\log_{14} 35 = a + b \Rightarrow 14^{a + b} = 35$. From here it follows that $(a + b) \log_{35} 14 = 1 \Rightarrow \log_{35} 14 = \frac{1}{a + b}$. From this you should be able to get an expression for $\log_{35} 28$ using a simple log rule.

3. Hello, ntrantrinh!

If the second one is base-10, I have a solution.

Let $\log 36 = a$ and $\log 125 = b$.

Find $\log\left(\tfrac{1}{12}\right)$ in terms of $a$ and $b$.

$\log(36) \,=\,a\quad\Rightarrow\quad \log(6^2) \,=\,a \quad\Rightarrow\quad 2\log(6) \,=\,a \quad\Rightarrow\quad \log(6) \,=\,\frac{a}{2}$

$\log(125) \,=\,b \quad\Rightarrow\quad \log(5^3)\,=\,b \quad\Rightarrow\quad 3\log(5) \,=\,b \quad\Rightarrow\quad \log(5) \,=\,\frac{b}{3}$

$\log\left(\frac{1}{12}\right) \;\;=\;\;\log\left(\frac{5}{60}\right) \;\;=\;\;\log(5) - \log(60) \;\;=\;\;\log(5) - \log(6\cdot10)$

. . . . $=\;\; \log(5) - \bigg[\log(6) + \log(10)\bigg] \;\;=\;\;\log(5) - \bigg[\log(6) + 1\bigg]$

. . . . $=\;\; \log(5) - \log(6) - 1 \;\;=\;\;\frac{b}{3} - \frac{a}{2} - 1$