Question

For a set of values of a, b if ${{\log }_{10}}2=a$ and ${{\log }_{10}}3=b$ then log 5.4 can be expressed as
A. a -3b-1
B. a+3b+1
C. a+3b-1
D. a-3b+1

Answer 1

To solve this question, we should know a property used in logarithms. That is ${{\log }_{x}}\left( \dfrac{a\times b\times c...}{d\times e..} \right)={{\log }_{x}}a+{{\log }_{x}}b+{{\log }_{x}}c....-{{\log }_{x}}d-{{\log }_{x}}e...$. To use this property in the question, we have to write 5.4 as a product of 2, 3, and 10. After writing 5.4 in terms of 2, 3, and 10, we apply logarithm to the equation and we use the property of logarithms to get the required answer.

Complete step-by-step solution:
In the question, it is given that ${{\log }_{10}}2=a$ and ${{\log }_{10}}3=b$ and we are asked to find the value of ${{\log }_{10}}5.4$. We have to do the factorisation of 5.4 and write it in terms of 2, 3, 10. We can do the factorisation of 54 and divide the factors by 10 to get the factors for 5.4. By doing prime factorisation of 54, we get
$\begin{aligned} & 2\left| \\!{\underline {\, 54 \,}} \right. \\\ & 3\left| \\!{\underline {\, 27 \,}} \right. \\\ & 3\left| \\!{\underline {\, 9 \,}} \right. \\\ & 3\left| \\!{\underline {\, 3 \,}} \right. \\\ & \left| \\!{\underline {\, 1 \,}} \right. \\\ \end{aligned}$
From this we can write, $54=2\times 3\times 3\times 3$
As we know that $5.4=\dfrac{54}{10}$, writing this in the above equation, we get
$5.4=\dfrac{54}{10}=\dfrac{2\times 3\times 3\times 3}{10}$
Applying logarithm of base 10 on both sides, we get
${{\log }_{10}}5.4={{\log }_{10}}\left( \dfrac{2\times 3\times 3\times 3}{10} \right)\to \left( 1 \right)$
We have to apply the property related to logarithms in the above equation. The property is
${{\log }_{x}}\left( \dfrac{a\times b\times c...}{d\times e..} \right)={{\log }_{x}}a+{{\log }_{x}}b+{{\log }_{x}}c....-{{\log }_{x}}d-{{\log }_{x}}e...$
Using this property in the equation-1, we get
${{\log }_{10}}5.4={{\log }_{10}}2+{{\log }_{10}}3+{{\log }_{10}}3+{{\log }_{10}}3-{{\log }_{10}}10$
We know that ${{\log }_{a}}a=1$.
Using this in the above equation and simplifying, we get
${{\log }_{10}}5.4={{\log }_{10}}2+3\times {{\log }_{10}}3-1$
In the question, it is given that ${{\log }_{10}}2=a$ and ${{\log }_{10}}3=b$.
Substituting a and b in above equation, we get
${{\log }_{10}}5.4=a+3b-1$
$\therefore {{\log }_{10}}5.4=a+3b-1$. The answer is option-C.

Note: The problem can be done in another way. That is
${{\log }_{x}}a=n\Rightarrow a={{x}^{n}}$
Using this relation, we can write
$\begin{aligned} & {{\log }_{10}}2=a\Rightarrow 2={{10}^{a}} \\\ & {{\log }_{10}}3=b\Rightarrow 3={{10}^{b}} \\\ \end{aligned}$
As we know

& {{\log }_{10}}5.4={{\log }_{10}}\left( \dfrac{2\times 3\times 3\times 3}{10} \right) \\\ & {{\log }_{10}}5.4={{\log }_{10}}\left( \dfrac{{{10}^{a}}\times {{10}^{b}}\times {{10}^{b}}\times {{10}^{b}}}{10} \right) \\\ \end{aligned}$$ $\dfrac{{{a}^{x}}\times {{a}^{y}}}{{{a}^{z}}}={{a}^{x+y-z}}$. Using this, we get $\begin{aligned} & {{\log }_{10}}5.4={{\log }_{10}}\left( \dfrac{{{10}^{a}}\times {{10}^{3b}}}{10} \right) \\\ & {{\log }_{10}}5.4={{\log }_{10}}\left( {{10}^{a+3b-1}} \right) \\\ \end{aligned}$ We have to use the property ${{\log }_{a}}{{a}^{n}}=n{{\log }_{a}}a=n$ ${{\log }_{10}}5.4={{\log }_{10}}\left( {{10}^{a+3b-1}} \right)=\left( a+3b-1 \right){{\log }_{a}}a=a+3b-1$ The answer is option-C.