Question

EDU 2020

Answer 1

Answer:

$\log_615=1.511$

$\log_{14}6-\log_{14}\frac{3}{2}=\log_{14}4$

$\log_w \frac{x}{z}=\log_wx-\log_w z$

Step-by-step explanation:

By quotient property :

$\log_xa-\log_xb=\log_x\frac{a}{b}$

Given:

$\log_630\approx 1.898$ and $\log_62\approx 0.387$

To find $\log_615$

Solution:

Applying quotient property:

$\log_630-\log_62=\log_6\frac{30}{2}=\log_615$

$\log_615=\log_630-\log_62$

$\log_615=1.898-0.387$

$\log_615=1.511$

To write $\log_{14}6-\log_{14}\frac{3}{2}$ as a single logarithm.

Solution:

Applying quotient property:

$\log_{14}6-\log_{14}\frac{3}{2}=\log_{14}\frac{6}{\frac{3}{2}}$

$\log_{14}6-\log_{14}\frac{3}{2}=\log_{14}(6\times \frac{2}{3})$

$\log_{14}6-\log_{14}\frac{3}{2}=\log_{14}4$

To expand $\log_w \frac{x}{z}$

Solution:

Applying quotient property:

$\log_w \frac{x}{z}=\log_wx-\log_w z$

Answer 2

Answer:

A

Step-by-step explanation: