Question

If 2x=3y=6z Then 1/x + 1/y +1/z = ?

Answer 1

Given:

• $2^x=3^y=6^{-z}$

To Find:

• $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = ?$      

Solution:

Let's assume

Since there is mistake in the given question. The value of z should be negative

    $2^x = 3^y = 6^{-z} = k$

After taking log both sides, we get

     $2 = k^ \frac{1}{x}$

     $3 = k^ \frac{1}{y}$

     $6 = k^ \frac{-1}{z}$

Since,

          2×3=6

putting the values

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

We know that when base is same power is added

So,

     ⇒       $k^{\frac{1}{x} + \frac{1}{y} } = k^\frac{-1}{z}$

    ⇒         $\frac{1}{x} + \frac{1}{y} = \frac{-1}{z}$

    ⇒         $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

Thus, The value of  $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$