Question

Given log630 ≈ 1.898 and log62 ≈ 0.387, log615 =

Answer 1

$\log_630\approx1.898\\\\\log_62\approx0.387\\\\\log_615=\log_6\dfrac{30}{2}=\log_630-\log_62\approx1.898-0.387=1.511\\\\Used:\\\\\log_ab-\log_ac=\log_a\dfrac{b}{c}$

Answer 2

Answer:

The value of $log_6{15}$ is 1.511.

Step-by-step explanation:

It is given that $log_6{30}\approx 1.898$ and $log_6{2}\approx 0.387$.

We have to find the value of $log_6{15}$.

It can be written as

$log_6{15}=log_6\frac{30}{2}$

Using property of lo, we get

$log_6{15}=log_630-log_62$          $log_m\frac{a}{b}=log_ma-log_mb$

Substitute $log_6{30}\approx 1.898$ and $log_6{2}\approx 0.387$ in the above equation.

$log_6{15}=1.898-0.387$

$log_6{15}=1.511$

Therefore the value of $log_6{15}$ is 1.511.