Question

How do you simplify ${\log _3}5 + {\log _3}2$ ?

Answer 1

Hint : When one term is raised to the power of another term, the function is called an exponential function, for example $a = {x^y}$ . The inverse of the exponential functions are called logarithm functions, the inverse of the function given in the example is $y = {\log _x}a$ that is a logarithm function. In the given problem we have logarithm to the base 3. So we need to apply a change of base formula to it. That is ${\log _b}a = \dfrac{{{{\log }_{10}}a}}{{{{\log }_{10}}b}}$ . Furthermore, by using logarithm laws we can solve this.

Complete step-by-step answer :
Given,
${\log _3}5 + {\log _3}2$
Now consider ${\log _3}5$ .
Applying the formula ${\log _b}a = \dfrac{{{{\log }_{10}}a}}{{{{\log }_{10}}b}}$ . we have,
${\log _3}5 = \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}3}}{\text{ }} - - - - (1)$
Now consider ${\log _3}2$
Applying the formula ${\log _b}a = \dfrac{{{{\log }_{10}}a}}{{{{\log }_{10}}b}}$ . we have,
${\log _3}2 = \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}}{\text{ }} - - - - (2)$
Now adding equation (1) and (2) we have,
${\log _3}5 + {\log _3}2 = \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}3}} + \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}}$
$= \dfrac{{{{\log }_{10}}5}}{{{{\log }_{10}}3}} + \dfrac{{{{\log }_{10}}2}}{{{{\log }_{10}}3}}$
Now taking LCM we have,
$= \dfrac{{{{\log }_{10}}5 + {{\log }_{10}}2}}{{{{\log }_{10}}3}}$
In the numerator we can apply product rule of logarithm, that is $\log (x.y) = \log (x) + \log (y)$
$= \dfrac{{{{\log }_{10}}\left( {5 \times 2} \right)}}{{{{\log }_{10}}3}}$
$= \dfrac{{{{\log }_{10}}\left( {10} \right)}}{{{{\log }_{10}}3}}$
We know ${\log _{10}}\left( {10} \right) = 1$ ,
$= \dfrac{1}{{{{\log }_{10}}3}}$
We know ${\log _{10}}3 = 0.4771$
$= \dfrac{1}{{0.4771}}$
$= 2.09$
After rounding off we have,
${\log _3}5 + {\log _3}2 = 2.1$
So, the correct answer is “ 2.1”.

Note : To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is $\log (x.y) = \log (x) + \log (y)$ . Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is $\log \left( {\dfrac{x}{y}} \right) = \log x - \log y$ . Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is $\log {x^a} = a\log x$ . These are the basic rules we use while solving a problem that involves logarithm function.