1. ## logs

use the properties of logarithims to simplify the logarithmic expression:

-log base 5 of (1/15)

-log base 2 of (4^2 * 3^4)

-log base of log base 5 of (1/250)

thanks!!

2. Hello, ss103!

I'm not sure what they mean by "simplify", but here goes . . .

Use the properties of logarithims to simplify the logarithmic expression:

$\displaystyle 1)\;\;\log_5\left(\tfrac{1}{15}\right)$

$\displaystyle \log_5\left(\frac{1}{15}\right) \;=\;\log_5(1) - \log_5(15)$

. . . . . . . $\displaystyle = \;0 - \log_5(3\cdot5)$

. . . . . . . $\displaystyle = \;-\bigg[\log_5(3) + \log_5(5)\bigg]$

. . . . . . . $\displaystyle = \;-\log_5(3) - \log_5(5)$

. . . . . . . $\displaystyle = \;-\log_5(3) - 1$

$\displaystyle (2)\;\;\log_2\left(4^2\cdot3^4\right)$

$\displaystyle \log_2\left(4^2\cdot3^4\right) \;=\;\log_2(4^2) + \log_2(3^4)$

. . . . . . . . $\displaystyle = \;2\log_2(4) + 4\log_2(3)$

. . . . . . . . $\displaystyle = \;2\log_2(2^2) + 4\log_2(3)$

. . . . . . . . $\displaystyle = \;2\!\cdot2\log_2(2) + 4\log_2(3)$

. . . . . . . . $\displaystyle = \;4 + 4\log_2(3)$

This one was not written clearly, but I'll take a guess.

$\displaystyle (3)\;\;\log_5\left(\tfrac{1}{250}\right)$

$\displaystyle \log_5\left(\frac{1}{250}\right) \;=\;\log_5(1) - \log_5(250)$

. . . . . . .$\displaystyle = \;0 - \log_5(2\cdot5^3)$

. . . . . . .$\displaystyle = \;-\bigg[\log_5(2) + \log_5(5^3)\bigg]$

. . . . . . .$\displaystyle = \;-\log_5(2) - 3\log_5(5)$

. . . . . . .$\displaystyle = \;-\log_5(2) - 3$