Question

prove that the value of
$log_{10}(3)$
lies between 1/3 and 1/2​

Answer 1

$\large\text{ \boxed{\bold{[Topic:\ Logarithms]}} }$

It is a basic query that requires properties of logarithms.

Let us prove what we need.

$\large\text{ \boxed{\bold{[Step\ 1.]}} }$

We have two equations that state the same meaning.

$\large\text{ \cdots\longrightarrow\boxed{a^{x}=b\iff x=\log_{a}{b}.} }$

If we take logarithm on the left equation, -

$\large\cdots\longrightarrow x\log{a}=\log{b}$

$\large\text{ \cdots\longrightarrow x=\dfrac{\log{b}}{\log{a}}. }$

We proved it.

$\large\text{ \cdots\longrightarrow\boxed{\log_{a}{b}=\dfrac{\log{b}}{\log{a}}} }$

$\large\text{ \boxed{\bold{[Step\ 2.]}} }$

Then, -

$\large\text{ \cdots\longrightarrow\log_{10}3=\dfrac{1}{\log_{3}{10}}. }$

Let's move on.

$\large\text{ \cdots\longrightarrow\dfrac{1}{\log_{3}{27}}<\dfrac{1}{\log_{3}{10}}<\dfrac{1}{\log_{3}9}. }$

$\large\text{ \cdots\longrightarrow\dfrac{1}{3}<\dfrac{1}{\log_{3}{10}}<\dfrac{1}{2}. }$

Finally, -

$\large\cdots\longrightarrow\dfrac{1}{3}<\log_{10}{3}<\dfrac{1}{2}.$