Question

3x = 4y = 12z, prove that: z = xy/x+y

Answer 1

answer
hope it helps...

Answer 2

Correct question is

If ${3}^{x} = {4}^{y} = {12}^{z}$then prove that $z = \frac{xy}{x + y}$

Answer:

Given,

${3}^{x} = {4}^{y} = {12}^{z}$

Let,

${3}^{x} = {4}^{y} = {12}^{z} = k$

where k is constant.

Now,

${3}^{x} = k \\ {3}^{ {x}^{ \frac{1}{x} } } = {k}^{ \frac{1}{x} } \\ {3}^{x \times \frac{1}{x} } = {k}^{ \frac{1}{x} } \\ 3 = {k}^{ \frac{1}{x} }$

and

${4}^{y} = k \\ {4}^{ {y}^{ \frac{1}{y} } } = {k}^{ \frac{1}{y} } \\ {4}^{y \times \frac{1}{y} } = {k}^{ \frac{1}{y} } \\ 4 = {k}^{ \frac{1}{y} }$

and

${12}^{z} = k \\ {12}^{ {z}^{ \frac{1}{z} } } = {k}^{ \frac{1}{z} } \\ {12}^{z \times \frac{1}{z} } = {k}^{ \frac{1}{z} } \\ 12 = {k}^{ \frac{1}{z} }$

We know,

$12 = 3 \times 4 \\ {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x} } \times {k}^{ \frac{1}{y} } \\ {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x} + \frac{1}{y} } \\ \frac{1}{z} = \frac{1}{x} + \frac{1}{y} \\ \frac{1}{z} = \frac{x + y}{xy } \\ z = \frac{xy}{x + y}$

[proved]

Here applied formulas are

${a}^{ {b}^{ \frac{1}{b} } } = {a}^{b \times \frac{1}{b} } = {a}^{1} = a$

$if \: {a}^{p} = {a}^{q} \: then \: p = q$