Question

Please help logarithms!

Answer 1

Given:

$\log_34\approx 1.262$

$\log_37\approx 1.771$

To find:

The value of $\log_3\left(\dfrac{4}{49}\right)$.

Solution:

We have,

$\log_34\approx 1.262$

$\log_37\approx 1.771$

Using properties of log, we get

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349$      $\left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]$

$\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2$      

$\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37$          $[\log x^n=n\log x]$

Substitute $\log_34\approx 1.262$ and $\log_37\approx 1.771$.

$\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)$

$\log_3\left(\dfrac{4}{49}\right)=1.262-3.542$

$\log_3\left(\dfrac{4}{49}\right)=-2.28$

Therefore, the value of $\log_3\left(\dfrac{4}{49}\right)$ is $-2.28$.