Question

If we are given the logarithmic equation ${\log }_{\sqrt{3}}5=a$ and ${\log }_{\sqrt{3}}2=b$ then ${\log }_{\sqrt{3}}300=$
$A)2(a+b)$
$B)2(a+b+1)$
$C)2(a+b+2)$
$D)a+b+4$
$E)a+b+1$

Answer 1

Hint: First, we will understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers and base of a number, to understand it better which is $\log x^m=m\log x$
It’s basically a problem based on the logarithmic properties. We will mainly use two properties of the logarithm to solve this problem.

Formula used:
> Using the logarithm law, $\log (ab)=\log a+\log b$
> ${\log }_{a^n}b={\displaystyle \frac{1}{n}}{\log }_{a}b$ where n is any number that we can choose as the base of the log

Complete step-by-step solution:
Given that we have two logarithm values with the same base $$\sqrt{3}$$ and even the required function needs the same base and so it is easy to solve with the log properties.
Since we have ${\log }_{\sqrt{3}}5=a$ and ${\log }_{\sqrt{3}}2=b$. By using these two functions we need to find the value of ${\log }_{\sqrt{3}}300$
Let us convert the log function, ${\log }_{\sqrt{3}}300={\log }_{\sqrt{3}}3+{\log }_{\sqrt{3}}100$ by using multiplication.
And now $100={10}^2$ then we have ${\log }_{\sqrt{3}}300={\log }_{\sqrt{3}}3+{\log }_{\sqrt{3}}{10}^2$
Now by use of the property $\log x^m=m\log x$, then we get ${\log }_{\sqrt{3}}300={\log }_{\sqrt{3}}3+2({\log }_{\sqrt{3}}10)$
Let $10=2\times 5,3={\sqrt{3}}^2$ (general algebraic expansion) substituting these values we get ${\log }_{\sqrt{3}}300={\log }_{\sqrt{3}}3+2({\log }_{\sqrt{3}}10)\Rightarrow {\log }_{\sqrt{3}}{\sqrt{3}}^2+2({\log }_{\sqrt{3}}2\times 5)$
Now using the properties $\log x^m=m\log x$ and $\log (ab)=\log a+\log b$ then we get ${\log }_{\sqrt{3}}300=2{\log }_{\sqrt{3}}\sqrt{3}+2({\log }_{\sqrt{3}}2+{\log }_{\sqrt{3}}5)$
Since in the logarithm, same base value and functions get one which is ${\log }_{a}a=1$ then we get ${\log }_{\sqrt{3}}300=2+2({\log }_{\sqrt{3}}2+{\log }_{\sqrt{3}}5)$ where ${\log }_{\sqrt{3}}\sqrt{3}=1$
From the given that we have ${\log }_{\sqrt{3}}5=a$ and ${\log }_{\sqrt{3}}2=b$ substituting these values we get ${\log }_{\sqrt{3}}300=2+2({\log }_{\sqrt{3}}2+{\log }_{\sqrt{3}}5)\Rightarrow 2+2(a+b)$
Thus, we get ${\log }_{\sqrt{3}}300=2(a+b+1)$
Therefore, the option $B)2(a+b+1)$ is correct.

Note: The other more general properties of the logarithm are
 ${\log }_{a^n}b={\displaystyle \frac{1}{n}}{\log }_{a}b$ where n is any number that we can choose as the base of the log
$\log ({\displaystyle \frac{a}{b}})=\log a-\log b$
The log rules can be used for the fast exponent calculation using the multiplication operations. The most general base of the log is $e$