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tankabanditka [31]
2 years ago
10

Please help logarithms!

Mathematics
1 answer:
nlexa [21]2 years ago
5 0

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.

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