Question

if 3.7^x=(0.37)^y=1000 show that x^-1-y^-1=3^-1

Answer 1

$x^{- 1}$ + y^{- 1} [/tex]= 3^{- 1}[/tex][/tex], it is proved.

Step-by-step explanation:

Given,

$3.7^{x} =[tex]0.37^{y} = 1000</p><p>Consider, [tex]3.7^{x}$ = 100            ....... (1)

Taking log on both sides in equation (1), we get

x·㏒3.7$[/tex] = 3·㏒10[tex]$

⇒ x·㏒3.7$[/tex] = 3</p><p>⇒ ㏒3.7[tex]$ = $\frac{3}{x}$     ....... (2)

Also,

Consider $0.37^{y} = 1000                   <strong>....... (3)</strong></p><p>Taking log on both sides in equation (3), we get</p><p>y·㏒0.37[tex]$ = 3·㏒10$[/tex] </p><p>⇒ y·㏒0.37[tex]$ = 3

⇒ ㏒0.37$[/tex] = [tex]\frac{3}{y}$   ....... (4)

Subtracting (4) into (2), we get

㏒3.7$[/tex]  - ㏒0.37[tex]$ = $\frac{3}{x}$  - $\frac{3}{y}$

⇒ ㏒10 $\frac{3}{x}$  - $\frac{3}{y}$

∴ $x^{- 1}$ + y^{- 1} [/tex]= 3^{- 1}[/tex][/tex], it is proved.