Question

4^x=0.008^y=10^z ...Find the relation between x,y,z

Answer 1

$\quad\quad \bold{\frac{1}{2x}=\frac{1}{3y}+\frac{1}{z}}$

Step-by-step explanation:

Let us take

$\quad\quad 4^{x}=0.008^{y}=10^{z}=k\:(\neq 0)$

Taking the first term, we get

$\quad 4^{x}=k$

$\to (2^{2})^{x}=k$

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

Taking the third term, we get

$\quad 10^{z}=k$

$\to 10=k^{\frac{1}{z}}\quad ...(ii)$

Taking the second term, we get

$\quad 0.008^{y}=k$

$\to \big(\frac{8}{1000}\big)^{y}=k$

$\to \big(\frac{2}{10}\big)^{3y}=k$

$\to \big(\frac{k^{\frac{1}{2x}}}{k^{\frac{1}{z}}}\big)^{3y}=k\quad [by\:(i),\:(ii)]$

$\to k^{(\frac{1}{2x}-\frac{1}{z})3y}=k$

Equating powers of k from both sides, we get

$\quad \big(\frac{1}{2x}-\frac{1}{z}\big)\:3y=1$

$\to \frac{1}{2x}-\frac{1}{z}=\frac{1}{3y}$

$\to \boxed{\frac{1}{2x}=\frac{1}{3y}+\frac{1}{z}}$

This is the required relation between x, y and z.

Related question:

If x = log(a + b), y = log(a - b), z = log(a² - b²). Then what is the relation between x, y, z? - https://brainly.in/question/13203506

Answer 2

Given as :

$4^{x}$  = $0.008^{y}$ = $10^{z}$

To Find :

The the relation between x , y , z

Solution :

$4^{x}$  = $0.008^{y}$ = $10^{z}$

$2^{2x}$ = $0.2^{3y}$ = $10^{z}$

Let   $2^{2x}$ = $0.2^{3y}$ = $10^{z}$  = k

So,  $2^{2x}$ = k

Or,  2 = $k^{\dfrac{1}{2x}}$       ...........1

Again

$0.2^{3y}$   = k

Or,  0.2 = $k^{\dfrac{1}{3y}}$         ..........2

Similarly

$10^{z}$  = k

Or,  10 = $k^{\dfrac{1}{z}}$          ......3

From, eq 1 and 3

 $\dfrac{2}{10}$  = $\dfrac{k^{\dfrac{1}{2x}}}{k^{\frac{1}{z}}}$

Or,  0.2 = $k(^{ \dfrac{1}{2x} -\dfrac{1}{z} })$      .........4

Now, From eq 2 and 4

$k(^{ \dfrac{1}{2x} -\dfrac{1}{z} })$   =   $k^{\dfrac{1}{3y}}$      

removing base k from both side

∴  $\dfrac{1}{2x}$ - $\dfrac{1}{z}$ = $\dfrac{1}{3y}$

Or,  $\dfrac{1}{2x}$  = $\dfrac{1}{3y}$ + $\dfrac{1}{z}$

Hence, The relation between x y  z is  $\dfrac{1}{2x}$  = $\dfrac{1}{3y}$ + $\dfrac{1}{z}$  Answer