• Feb 25th 2010, 03:41 AM
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) =
• Feb 25th 2010, 04:08 AM
Quote:

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 $\log_{5}(\sqrt{\frac{x-1}{x+1}})$ ?

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

$
=\log_{2}7
$
• Feb 25th 2010, 04:12 AM
aligator0207
Yes, that's how you write out number 1. Do you know how to do number 2?
• Feb 25th 2010, 04:19 AM
Soroban
Hello, aligator0207!

Quote:

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

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

Quote:

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

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

Quote:

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

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

• Feb 25th 2010, 04:21 AM
aligator0207
Sweet! Thank you so much. At least I got the first two right (Rock)