Question

2.3^x + 3.2^y =12, 3^x=2^y +1

Answer 1

Given that,

2.3^x=0.23^y=1000

Now,

2.3^x=1000 and 0.23^y=1000

Applying log on both sides we get,

log2.3^x=log1000

xlog2.3=log10^3

xlog(23/10)=3

log(23/10)=3/x.....................................1

and,

log0.23^y=log1000

ylog(23/100)=log10^3

ylog(23/100)=3

log(23/100)=3/y..................................2

By eq1-eq2 we get,

log(23/10)-log(23/100)=3/x-3/y

log{(23/10)/(23/100}=3(1/x-1/y)

log10=3(1/x-1/y)

1=3(1/x-1/y)

1/3=1/x-1/y.

Hence value of 1/x-1/y=1/3.