# How do you evaluate \log _ { 7} 3+ \frac { \log _ { 7} 5} { 2} + \frac { \log _ { 7} 11} { 2}?

May 20, 2017

${\log}_{7} \sqrt{495}$

$\approx {\log}_{7} 22.25$

$\approx 1.59$

#### Explanation:

Before beginning, you must know the following properties related to logarithms:-

1. $a \cdot {\log}_{c} b = {\log}_{c} \left({b}^{a}\right)$
2. ${\log}_{c} a + {\log}_{c} b = {\log}_{c} \left(a \cdot b\right)$

Now, in this problem,

${\log}_{7} 3 + {\log}_{7} \frac{5}{2} + {\log}_{7} \frac{11}{2}$

$= \frac{2 \cdot {\log}_{7} 3 + {\log}_{7} 5 + {\log}_{7} 11}{2}$

$= \frac{{\log}_{7} {3}^{2} + {\log}_{7} \left(5 \cdot 11\right)}{2}$

$= \frac{{\log}_{7} 9 + {\log}_{7} 55}{2}$

$= \frac{{\log}_{7} \left(9 \cdot 55\right)}{2}$

$= \frac{1}{2} \cdot {\log}_{7} 495$

$= {\log}_{7} \left({495}^{\frac{1}{2}}\right)$

$= {\log}_{7} \sqrt{495}$

$\approx {\log}_{7} 22.25$

$\approx 1.59$