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
Last updated date: 19th Mar 2023
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Answer
304.8k+ views
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.
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