Question

If 4.2)^x =(0.42)^y=100 then 1/x-1/y

Answer 1

Question:

$If \: {(4.2)}^{x} = {(0.42)}^{y} = 100 \: then \: \frac{1}{x} - \frac{1}{y}$

Given data:

${(4.2)}^{x} = {(0.42)}^{y} = 100$

To find:

$To \: find \: the \: value \: of \\ \frac{1}{x} - \frac{1}{y}$

Answer:

$\underline{ \: \bold{ \frac{1}{2} } \: }\:is\:the\:value.$

Step-by-step explanation:

${(4.2)}^{x} = {(0.42)}^{y} = 100 = {10}^{2}$

$\implies {(4.2)}^{x} = {10}^{2}$

Now considering log on both sides,

$log {(4.2)}^{x} = log{(10)}^{2}$

$x.log 4.2 = 2.log 10$

$We \: know \: that, \: \boxed{log 10 = 1}$

$x.log 4.2 = 2$

$x = \frac{2}{log 4.2}$

$\boxed{ \frac{1}{x} = \frac{log 4.2}{2} }$

$\implies {(0.42)}^{y} = {10}^{2}$

Now considering log on both sides,

$log {(0.42)}^{y} = {(10)}^{2}$

$y.log 0.42 = 2.log 10$

$We \: know \: that, \: \boxed{log 10 = 1}$

$y.log 0.42 = 2$

$y = \frac{2}{log 0.42}$

$\boxed{ \frac{1}{y} = \frac{log 0.42}{2} }$

$\implies \frac{1}{x} - \frac{1}{y}$

$= \frac{log 4.2}{2} - \frac{0.42}{2}$

$= \frac{log \frac{4.2}{0.42} }{2}$

$= \frac{log 10}{2}$

$= \boxed{ \frac{1}{2} }$

$Here\:log 10 = 1$

Answer 2

If 4.2)^x =(0.42)^y=100 then 1/x-1/y

0.5 = 1/2

is the answer