Question

If $\log_{10}3=0.4771212$, without using log tales, find $\log_{10}9, \ \log_{10}\sqrt{3}, \ \log_{10}\frac{1}{9} \ and \ \log_{10}(0.3)$.

Answer 1

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Answer 2

Answer:

(i) $log_{10}9=0.954$

(ii) $log_{10}\sqrt{3}=0.2385$

(iii) $log_{10}\frac{1}{9}=-0.954$

(iv) $log_{10}0.3=-0.523$

Step-by-step explanation:

In the given question,

We have the value given to us of the term,

$log_{10}3=0.4771212$

So, we have to find the value of the other terms from the given term's value.

Now, from the properties of the logarithms we have,

$loga^{n}=nloga\\and,\\log\frac{a}{b}=loga-logb$

So, doing the same we get,

i) $log_{10}9=log_{10}(3)^{2}=2log_{10}3=2\times 0.4771212=0.954$

Therefore,

$log_{10}9=0.954$

ii) $log_{10}\sqrt{3}=\frac{1}{2}log_{10}3=\frac{1}{2}(0.4771212)=0.2385$

Therefore,

$log_{10}\sqrt{3}=0.2385$

iii) $log_{10}\frac{1}{9}=log_{10}1-log_{10}(9)=-log_{10}9=-0.954$

Therefore,

$log_{10}\frac{1}{9}=-0.954$

iv) $log_{10}0.3=log_{10}\frac{3}{10}=log_{10}3-log_{10}10=log_{10}3-1=0.477-1=-0.523$

Therefore,

$log_{10}0.3=-0.523$

Therefore,

We have all the values of the required terms as 0.954, 0.2385, -0.954 and -0.523 respectively.