Question

if 3^x=5^y=75^z, show that z=xy/2x+y

Answer 1

Answer:

Step-by-step explanation:

Let $3^{x}=5^{y}=75^{z}= M$, then

$3= M^{\frac{1}{x}}$, $5= M^{\frac{1}{y}}$ and $75= M^{\frac{1}{z}}$.

Also, 75 can be written as: $75=5^{2}{\times}3$

$M^{\frac{1}{z}}= M^{\frac{2}{y}}{\times}M^{\frac{1}{x}}$

$M^{\frac{1}{z}}=M^{\frac{2}{y}+\frac{1}{x}}$

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

$\frac{xy}{z}=2x+y$

$z=\frac{xy}{2+y}$

Hence proved.

Answer 2

Answer with Step-by-step explanation:

Since we have given that

$3^x=5^y=75^z$

Consider,

$3^x=5^y=75^z=k\\\\3=k^{\frac{1}{x}}\\\\5=k^{\frac{1}{y}}\\\\75=k^{\frac{1}{z}$

According to question,

$75=5^2\times 3\\\\k^{\frac{1}{z}}=k^{\frac{2}{y}}\times k^{\frac{1}{x}}\\\\\dfrac{1}{z}=\dfrac{2}{y}+\dfrac{1}{x}\\\\\dfrac{1}{z}=\dfrac{2x+y}{xy}\\\\z=\dfrac{xy}{2x+y}$

Hence, proved.