Question

Please solve this question. Only a genius can do it .

Answer 1

$\underline{\underline{\bold{Question:}}}$

$\tt{If\quad (2.381)^x=(0.2381)^y=(10)^z\quad ,\:then\:\:the\:\:value\:\:of}$

$\tt{\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x}=?}$

$\bold{Solution:}$

$\tt{Let\quad(2.381)^x=(0.2381)^y=(10)^z\:be\:k.}\\\\\\\tt{So}\\\\\\\implies{\tt{(2.381)^x=k}}\\\\\therefore\quad{\tt{2.381=k^{\frac{1}{x}}}}\\\\\\\implies{\tt{(0.2381)^y=k}}\\\\\therefore\quad 0.2381=k^{\frac{1}{y}}\\\\\\\\\implies{\tt{10^z=k}}\\\\\therefore\quad 10=k^{\frac{1}{z}}$

From above conditions.

$\implies{\bold{\dfrac{2.381}{0.2381}=10}}\\\\\\\underline{\bold{Put\:the\:value,}}\\\\\\\implies{\bold{\dfrac{k^{\frac{1}{x}}}{k^{\frac{1}{y}}}=k^{\frac{1}{z}}}}\\\\\\\implies{\bold{k^{(\frac{1}{x}-\frac{1}{y})}=k^{\frac{1}{z}}}}\\\\\\\implies{\bold{\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{z}}}\\\\\\\implies{\bold{\dfrac{1}{z}+\dfrac{1}{y}-\dfrac{1}{x}=0}}$

$\boxed{\boxed{\bold{\dfrac{1}{z}+\dfrac{1}{y}-\dfrac{1}{x}=0}}}$

Answer 2

$\mathfrak{\huge{Answer:}}$

Given that:

$\tt{2.381^x = 0.2381^{y} = 10^z}$

Let all these values be = k

In that way, we can assume :

$\tt{2.381^x = 0.2381^{y} = 10^z = k}$

Now, individually, the values of all these terms will be :

$\sf{2.381 = k^\frac{1}{x}}$

And :

$\sf{0.2381 = k^\frac{1}{y}}$

The last term will be :

$\sf{10 = k^\frac{1}{z}}$

In these types of questions, after simplifying the values up to this extent, we need to find relations between the numeric values. Here, we can state the relation :

$\sf{0.2381 \times 10 = 2.381}$

Put the variables in place of these numeric values :

$\sf{k^\frac{1}{y} \times k^\frac{1}{z} = k^\frac{1}{x}}$

Using the property of exponents, we can write :

$\tt{\frac{1}{y} + \frac{1}{z} = \frac{1}{x}}$

=》 $\tt{\frac{1}{y} + \frac{1}{z} - \frac{1}{x} = 0}$

That's your answer.

Final answer : $\boxed{\bold{(3)\:0}}$