Question

3^x = 5^y = 75^z proof that xz/x+y​

Answer 1

Correct Question:

• If $\sf 3^{x} = 5^{y} = 75^{z}$, prove that, $\sf z = \dfrac{xy}{2x+y}$

Proof:

$\sf Let \ {3}^{x} = {5}^{y} = {75}^{z} = k$

So,

$\sf \mapsto{3}^{x} = k \implies 3 = {k}^{ \frac{1}{x} } \: ...(i)$

$\sf \mapsto{5}^{y} = k \implies 5 = {k}^{ \frac{1}{y} } \: ...(ii)$

$\sf \mapsto{75}^{z} = k \implies 75 = {k}^{ \frac{1}{z} } \: ...(iii)$

Again,

$\sf \mapsto 75 = 3 \times {5}^{2}$

Substituting the values, we get,

$\sf \mapsto {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x} } \times ( {k}^{ \frac{1}{y} })^{2}$

$\sf \mapsto {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x} } \times {k}^{ \frac{2}{y} }$

$\sf \mapsto {k}^{ \frac{1}{z} } = {k}^{ \frac{1}{x} + \frac{2}{y} }$

Comparing base, we get,

$\sf \mapsto \small \dfrac{1}{z} = \dfrac{1}{x} + \dfrac{2}{y}$

$\sf \mapsto \small \dfrac{1}{z} = \dfrac{y + 2x}{xy}$

$\sf \mapsto \small \dfrac{1}{z} = \dfrac{2x + y}{xy}$

$\sf \mapsto z = \dfrac{xy}{2x + y}$

(Hence Proved)

Formulae to solve indices questions:

$\sf \mapsto {x}^{0} = 1,x \neq0$

$\sf \mapsto {x}^{a} \times {x}^{b} = {x}^{a + b}$

$\sf \mapsto({x}^{a} )^{b} = {x}^{ab}$

$\sf \mapsto \dfrac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b} ,x \neq0$

$\sf \mapsto {x}^{a} \times {y}^{a} = {(xy)}^{a}$

$\sf \mapsto \bigg(\dfrac{x}{y} \bigg)^{a} = \dfrac{ {x}^{a} }{ {y}^{a} } ,y \neq0$

$\sf \mapsto {x}^{a} = {x}^{b} \implies a = b,x \neq0$

$\sf \mapsto {x}^{a} = {y}^{a} \implies x = y,a \neq0$

$\sf \mapsto {x}^{a} = \dfrac{1}{ {x}^{ - a} }$

$\sf \mapsto \sqrt[y]{x} = {x}^{ \frac{1}{y} }$