Question

How do you expand log $ A{{B}^{2}} $ ?

Answer 1

Hint: Hint: We have to expand $ \log \left( A{{B}^{2}} \right) $ , to do so we will learn the properties of log,
We will use that $ \log \left( xy \right) $ is written as $ \log \left( x \right)+\log \left( y \right) $ .
We will use $ \log \left( \dfrac{x}{y} \right)=\log \left( x \right)-\log \left( y \right) $
To simplify we will also use $ \log \left( {{x}^{y}} \right)=y\log \left( x \right) $ . We will learn that exponential and log are both connected functions we will use above properties to expand $ \log \left( A{{B}^{2}} \right) $ .

Complete step by step answer:
We are asked to expand $ \log \left( A{{B}^{2}} \right) $ .
To solve the problem, we will learn about how the log function behaves. We will first understand the log function, learn its various properties.
Before we start the log, we should know that log is just the inverse function of the exponential function. It behaves most similar to that exponential function.
Thus we know that when we multiply two terms with the same base then their power gets added up.
 $ {{x}^{a}}\times {{x}^{b}}={{x}^{a+b}} $
Similarly, when we multiply two terms in log function then value of the log of those terms added up, that is –
 $ \log \left( xy \right)=\log \left( x \right)+\log \left( y \right) $ …………………….. (1)
As we know that when we divide two term with same base then their power get subtracted that is $ \dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}} $
Similarly, when we apply log on $ \dfrac{x}{y} $ then their value is subtracted.
That is –
 $ \log \left( \dfrac{x}{y} \right)=\log x-\log y $
We also know that –
 $ {{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}} $
So, similarly in log, $ \log \left( {{x}^{y}} \right)=y\log x $ …………………………………….. (2)
We also know that –
 $ \log \left( 1 \right)=0 $
So, we will use these properties to find the expanded term of our given problem.
Now, we have –
 $ \log \left( A{{B}^{2}} \right) $
Let us consider $ A=x $ and $ {{B}^{2}}=y $ .So,
 $ \log \left( A{{B}^{2}} \right)=\log \left( xy \right) $
Now, using $ \log \left( xy \right)=\log \left( x \right)+\log \left( y \right) $ we get –
 $ \log \left( A{{B}^{2}} \right)=\log \left( A \right)+\log \left( {{B}^{2}} \right) $ ………………………………. (3)
Now we use $ \log \left( {{x}^{y}} \right)=y\log x $ on $ \log \left( {{B}^{2}} \right) $

We get –
 $ \log \left( {{B}^{2}} \right)=2\log \left( B \right) $
Using this in equation (3), we get –
 $ \log \left( A{{B}^{2}} \right)=\log \left( A \right)+2\log \left( B \right) $
So, we get –
Expand $ \log \left( A{{B}^{2}} \right) $ as $ \log \left( A{{B}^{2}} \right)=\log \left( A \right)+2\log \left( B \right) $.

Note:
Remember to solve such problem we had to follow the properties is we using in appropriate properties like $ \log \left( xy \right)=\log \left( x \right)\times \log \left( y \right) $ and $ \log \left( \dfrac{x}{y} \right)=\dfrac{\log \left( x \right)}{\log \left( y \right)} $ this we will reach to a wrong answer. We will simplify step by step so that it is clear so that we get no error in the way.