Question

Given that $a,b\in \left\{ 0,1,2,.....9 \right\}$ with $a+b\ne 0$ and ${{\left( a+\dfrac{b}{10} \right)}^{x}}={{\left( \dfrac{a}{10}+\dfrac{b}{100} \right)}^{y}}=1000$ .Then $\left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=$
1) $1$
2) $\dfrac{1}{2}$
3) $\dfrac{1}{3}$
4) $\dfrac{1}{4}$

Answer 1

Hint: Here we have been given two exponent terms equal to a number and we have to find the value of the variable given. Firstly we will take both the elements separately equal to the number and find the value of both the variables by taking the logarithm function on both sides. Then we will substitute the value in the term we have to find and simplify it to get the desired answer.

Complete answer: The exponent are given as follows,
${{\left( a+\dfrac{b}{10} \right)}^{x}}={{\left( \dfrac{a}{10}+\dfrac{b}{100} \right)}^{y}}=1000$……$\left( 1 \right)$
Where, $a,b\in \left\{ 0,1,2,.....9 \right\}$ with $a+b\ne 0$
We have to find the value of,
$\left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)$…..$\left( 2 \right)$
Now firstly from equation (1) we get,
${{\left( a+\dfrac{b}{10} \right)}^{x}}=1000$
Taking $\log $ both sides we get,
$\log {{\left( a+\dfrac{b}{10} \right)}^{x}}=\log 1000$
Now as we know $\log {{a}^{b}}=b\log a$ using it above we get,
$\Rightarrow x\log \left( a+\dfrac{b}{10} \right)=\log 1000$
$\Rightarrow x=\dfrac{\log 1000}{\log \left( a+\dfrac{b}{10} \right)}$….$\left( 3 \right)$
Next from equation (1) we get,
${{\left( \dfrac{a}{10}+\dfrac{b}{100} \right)}^{y}}=1000$
Taking $\log $ both sides we get,
$\log {{\left( \dfrac{a}{10}+\dfrac{b}{100} \right)}^{y}}=\log 1000$
Now as we know $\log {{a}^{b}}=b\log a$ using it above we get,
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=y\log \left( \dfrac{a}{10}+\dfrac{b}{100} \right)=\log 1000$
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=y=\dfrac{\log 1000}{\log \left( \dfrac{a}{10}+\dfrac{b}{100} \right)}$…..$\left( 4 \right)$
Substitute the values from equation (3) and (4) in equation (2) we get,
 $\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{1}{\dfrac{\log 1000}{\log \left( a+\dfrac{b}{10} \right)}}-\dfrac{1}{\dfrac{\log 1000}{\log \left( \dfrac{a}{10}+\dfrac{b}{100} \right)}}$
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{\log \left( a+\dfrac{b}{10} \right)}{\log 1000}-\dfrac{\log \left( \dfrac{a}{10}+\dfrac{b}{100} \right)}{\log 1000}$
As the denominator is same of both the values we get,
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{\log \left( a+\dfrac{b}{10} \right)-\log \left( \dfrac{a}{10}+\dfrac{b}{100} \right)}{\log 1000}$
 $\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{\log \left( \dfrac{10a+b}{10} \right)-\log \left( \dfrac{10a+b}{100} \right)}{\log 1000}$
Now as we know the property of logarithm that $\log a-\log b=\log \dfrac{a}{b}$ using it above we get,
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{\log \left( \dfrac{\dfrac{10a+b}{10}}{\dfrac{10a+b}{100}} \right)}{\log 1000}$
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{\log \left( \dfrac{10a+b}{10}\times \dfrac{100}{10a+b} \right)}{\log 1000}$
On simplifying further we get,
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{\log 10}{\log 1000}$
Using the calculator we get the value of $\log 10=1$ and $\log 1000=3$ substituting them above we get,
$\Rightarrow \left( \dfrac{1}{x} \right)-\left( \dfrac{1}{y} \right)=\dfrac{1}{3}$
Hence the correct option is (3).

Note:
This type of question looks difficult when we see them but once we start solving them they get easy and predictable as we proceed to the next step. As we want the value of the unknown variable we have to remove them from the exponent and for doing that using logarithm is the only way also different properties of logarithm should be known in order to solve such questions.