Expanding:

Combining the expression:

Log(base 5)(x^2 - 1) - Log(base 5)(x - 1) =

Simplify:

(Log(base 2)5)(Log(base 5)7) =

2. Originally Posted by aligator0207
Expanding:

Combining the expression:

Log(base 5)(x^2 - 1) - Log(base 5)(x - 1) =

Simplify:

(Log(base 2)5)(Log(base 5)7) =
hi

for (1) , is it $\displaystyle \log_{5}(\sqrt{\frac{x-1}{x+1}})$ ?

(2) $\displaystyle \log_{2}5\cdot \log_{5}7=\frac{\log_{2}5}{\log_{2}2}\cdot \frac{\log_{2}7}{\log_{2}5}$

$\displaystyle =\log_{2}7$

3. Yes, that's how you write out number 1. Do you know how to do number 2?

4. Hello, aligator0207!

Expand: .$\displaystyle \log_5 \sqrt{\frac{x-1}{x+1}}$

$\displaystyle \log_5\sqrt{\frac{x-1}{x+1}} \;=\;\log_5\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} \;=\;\tfrac{1}{2}\log_5\left(\frac{x-1}{x+1}\right) \;=\;\tfrac{1}{2}\bigg[\log_5(x-1) - \log_5(x+1)\bigg]$

Combine: .$\displaystyle \log_5(x^2 - 1) - \log_5(x - 1)$

$\displaystyle \log_5(x^2-1) - \log_5(x-1) \;=\;\log_5\left(\frac{x^2-1}{x-1}\right) \;=\;\log_5\left(\frac{(x-1)(x+1)}{x-1}\right) \;=\;\log_5(x+1)$

Simplify: .$\displaystyle \log_2(5)\cdot\log_5(7)$
Base-change Formula: .$\displaystyle \log_b(a) \:=\:\frac{\log (a)}{\log(b)}$

$\displaystyle \log_2(5)\cdot\log_5(7) \;=\;\frac{\log(5)}{\log(2)} \cdot\frac{\log(7)}{\log(5)} \;=\;\frac{\log(7)}{\log(2)} \;=\;\log_2(7)$

5. Sweet! Thank you so much. At least I got the first two right