Question

How do you condense $In2 + In6 - \left( {\dfrac{1}{2}} \right)In9?$

Answer 1

As per the given question we can solve it by using the logarithm formula. The formulae that we have to apply to solve the above given question is as follows: $Ina + Inb = In(a \times b)$ and there is also another formula that we can use is $Ina - Inb = In\left( {\dfrac{a}{b}} \right)$. These are the two basic formulas of logarithms and we can use it and then we can condense the given question.

Complete step by step solution:
From the given question we have to condense the equation is $In2 + In6 - \left( {\dfrac{1}{2}} \right)In9$.
As we have already discussed the basic logarithm formula that we have to use to condense the given equation. First we will use the formula i.e. $Ina + Inb = In(a \times b)$. By applying this formula in the first part of the equation we have: $In(2 \times 6) - \left( {\dfrac{1}{2}} \right)In9$.
Now we have to apply another formula to condense the above given question. We will use this formula now i.e. $\dfrac{1}{n}Inx = In(\sqrt[n]{x})$.
So by applying this $\dfrac{1}{2}In9 = In(\sqrt[{}]{9})$. Now by substituting all the values in the equation we have,
$In12 - In(\sqrt 9 ) = In12 - In3$. Here we will apply another basic logarithm formula mentioned above, So $In12 - In3 = In\left( {\dfrac{{12}}{3}} \right)$. On further simplifying we have $In4$.
Hence the required condensed equation is $In4$.

Note: Before solving this kind of question we should be fully aware of the logarithms rule and their formulas. Their basic logarithm properties and their basic formulae are very useful in solving questions. Behind these three there are many other logarithm formulae which can be used in the questions and we can efficiently use these formulae in solving the questions.