Logarithms. Please help.

**aligator0207** · February 25th 2010, 03:41 AM

Expanding:

Log(base 5)( radical x-1/x+1 )=

Combining the expression:

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

Simplify:

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

**mathaddict** · February 25th 2010, 04:08 AM

Originally Posted by aligator0207

Expanding:

Log(base 5)( radical x-1/x+1 )=

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

$<br /> =\log_{2}7<br />$

**aligator0207** · February 25th 2010, 04:12 AM

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

**Soroban** · February 25th 2010, 04:19 AM

Hello, aligator0207!

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

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

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

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

Simplify: . $\log_2(5)\cdot\log_5(7)$

Base-change Formula: . $\log_b(a) \:=\:\frac{\log (a)}{\log(b)}$

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

**aligator0207** · February 25th 2010, 04:21 AM

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

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