# If ${{\log }_{12}}18=\alpha $ and ${{\log }_{24}}54=\beta $, then the value of $\alpha \beta +\left( \alpha -\beta \right)$ will be:

(a) 2

(b) ${{\log }_{12}}24$

(c) 1

(d) none of these

Answer

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Hint: Use Base change rule to convert ${{\log }_{24}}54$ and ${{\log }_{12}}18$ into common $\log $ that is log to the base 10. Also, break 18 and 54 into their prime factors and use the product rule of log to simplify the value. Use ${{\log }_{10}}3\approx 0.48$ and ${{\log }_{10}}2\approx 0.3$, and substitute these values in the simplified equation of $\alpha $ and $\beta $.

Some important formulas for logarithms are:

$\begin{align}

& {{\log }_{m}}{{n}^{a}}=a{{\log }_{m}}n,\text{ } \\

& {{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n\text{ } \\

& \text{lo}{{\text{g}}_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n \\

& {{\log }_{{{a}^{b}}}}m=\dfrac{1}{b}{{\log }_{a}}m \\

\end{align}$

Complete step-by-step answer:

Now,$\alpha =$${{\log }_{12}}18=\dfrac{{{\log }_{10}}18}{{{\log }_{10}}12}$$=\dfrac{{{\log }_{10}}({{3}^{2}}\times 2)}{{{\log }_{10}}({{2}^{2}}\times 3)}=\dfrac{2\log 3+\log 2}{2\log 2+\log 3}$

and, $\beta ={{\log }_{24}}54=\dfrac{{{\log }_{10}}54}{{{\log }_{10}}24}=\dfrac{{{\log }_{10}}({{3}^{3}}\times 2)}{{{\log }_{10}}({{2}^{3}}\times 3)}=\dfrac{3{{\log }_{10}}3+{{\log }_{10}}2}{3{{\log }_{10}}2+{{\log }_{10}}3}$

Substituting the value of ${{\log }_{10}}2$ and ${{\log }_{10}}3$ in the expression of $\alpha \text{ and }\beta $, we get,

$\alpha =\dfrac{2\times 0.48+0.3}{2\times 0.3+0.48}\approx 1.167$

$\beta =\dfrac{3\times 0.48+0.3}{3\times 0.3+0.48}\approx 1.261$

Now, the given expression

$\begin{align}

& =\alpha \beta +(\alpha -\beta ) \\

& =(1.167\times 1.261)+(1.167-1.261) \\

& =1.471-0.094 \\

& =1.377 \\

\end{align}$

Now, calculate, ${{\log }_{12}}24=\dfrac{{{\log }_{10}}24}{{{\log }_{10}}12}=\dfrac{3{{\log }_{10}}2+{{\log }_{10}}3}{2{{\log }_{10}}2+{{\log }_{10}}3}=1.279$

Hence, we see that no option is correct. Therefore, the correct option is (d).

Note: Generally we have to remember the values of common logarithm having argument 1 to 10. Here, $\alpha $ and $\beta $ cannot be further simplified, so, we have to substitute the value of ${{\log }_{10}}2$ and ${{\log }_{10}}3$ in the expression to get the answer.

Some important formulas for logarithms are:

$\begin{align}

& {{\log }_{m}}{{n}^{a}}=a{{\log }_{m}}n,\text{ } \\

& {{\log }_{a}}\left( m\times n \right)={{\log }_{a}}m+{{\log }_{a}}n\text{ } \\

& \text{lo}{{\text{g}}_{a}}\left( \dfrac{m}{n} \right)={{\log }_{a}}m-{{\log }_{a}}n \\

& {{\log }_{{{a}^{b}}}}m=\dfrac{1}{b}{{\log }_{a}}m \\

\end{align}$

Complete step-by-step answer:

Now,$\alpha =$${{\log }_{12}}18=\dfrac{{{\log }_{10}}18}{{{\log }_{10}}12}$$=\dfrac{{{\log }_{10}}({{3}^{2}}\times 2)}{{{\log }_{10}}({{2}^{2}}\times 3)}=\dfrac{2\log 3+\log 2}{2\log 2+\log 3}$

and, $\beta ={{\log }_{24}}54=\dfrac{{{\log }_{10}}54}{{{\log }_{10}}24}=\dfrac{{{\log }_{10}}({{3}^{3}}\times 2)}{{{\log }_{10}}({{2}^{3}}\times 3)}=\dfrac{3{{\log }_{10}}3+{{\log }_{10}}2}{3{{\log }_{10}}2+{{\log }_{10}}3}$

Substituting the value of ${{\log }_{10}}2$ and ${{\log }_{10}}3$ in the expression of $\alpha \text{ and }\beta $, we get,

$\alpha =\dfrac{2\times 0.48+0.3}{2\times 0.3+0.48}\approx 1.167$

$\beta =\dfrac{3\times 0.48+0.3}{3\times 0.3+0.48}\approx 1.261$

Now, the given expression

$\begin{align}

& =\alpha \beta +(\alpha -\beta ) \\

& =(1.167\times 1.261)+(1.167-1.261) \\

& =1.471-0.094 \\

& =1.377 \\

\end{align}$

Now, calculate, ${{\log }_{12}}24=\dfrac{{{\log }_{10}}24}{{{\log }_{10}}12}=\dfrac{3{{\log }_{10}}2+{{\log }_{10}}3}{2{{\log }_{10}}2+{{\log }_{10}}3}=1.279$

Hence, we see that no option is correct. Therefore, the correct option is (d).

Note: Generally we have to remember the values of common logarithm having argument 1 to 10. Here, $\alpha $ and $\beta $ cannot be further simplified, so, we have to substitute the value of ${{\log }_{10}}2$ and ${{\log }_{10}}3$ in the expression to get the answer.

Last updated date: 01st Oct 2023

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