Question

7. If 2x = 3y = 6z, show that z =
xy/x+y​

Answer 1

Answer:

• Given: 2x = 3y = 6z

• To prove:  

Now,

Let 2x = 3y = 6z = k (where k is any constant)…(1)

From equation 1, we get

…(2)

…(4)

Multiplying equation 2 and 3 we get

(∵ aman = am + n)  

But, From(4)

Substituting  in the above equation, we get  

Hence Proved.

Answer 2

Answer

• z = xy / ( x + y )

Given

$\bullet\ \; \rm 2^x=3^y=6^z$

To Prove

$\bullet \; \; \rm z=\dfrac{xy}{x+y}$

Proof

Let ,

$\rm 2^x=3^y=6^z=k$

where , k is any constant

So ,

$\to \rm 2^x=k\\\\\to \rm 2=k^{\frac{1}{x}}\\\\\to \rm k^{\frac{1}{x} }=2...(1)$

and

$\to \rm 3^y=k\\\\\to \rm k^{\frac{1}{y}}=3...(2)$

and

$\to \rm 6^z=k\\\\\to \rm k^{\frac{1}{z}}=6...(3)$

Now , Multiply eq (1) & eq (2) ,

$\to \rm k^{\frac{1}{x}}\times k^{\frac{1}{y}}=2\times 3\\\\ \bf \because\ a^m\times a^n=a^{m+n}\\\\\to \rm k^{\frac{1}{x}+\frac{1}{y}}=6\\\\ \bf \because\ From\ (3)\\\\\to \rm k^{\frac{1}{x}+\frac{1}{y}}=k^{\frac{1}{z}}$

Since , bases are equal , so exponents must be equal .

$\to \rm \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{z}\\\\\to \rm \dfrac{y+x}{xy}=\dfrac{1}{z}\\\\\to \bf z=\dfrac{xy}{x+y}\ \; \pink{\bigstar}$

Hence proved .