The logarithm of both sides is based on 10. Let lg2=t, then lg5 =1-t.

alg2+blg5= 1

at+b( 1-t)= 1

(a- 1)t+t+(b- 1)( 1-t)+( 1-t)= 1

(a- 1)t+(b- 1)( 1-t)= 0

(a- 1)/(b- 1)= 1- 1/t

Similarly, 2 c * 5 d = 10.

(c- 1)/(d- 1)= 1- 1/t

So (a-1)/(b-1) = (c-1)/(d-1)

(a- 1)(d- 1)=(b- 1)(c- 1)

2.f(x)=(a^x-a^-x)/(a^x+a^-x)=(a^2x- 1)/(a^2x+ 1)

f(x)+f(y)=(a^2x- 1)/(a^2x+ 1)+(a^2y- 1)/(a^2y+ 1)

=[a^2(x+y)+a^2x-a^2y- 1+a^2(x+y)-a^2x+a^2y- 1]/[(a^2x+ 1)(a^2y+ 1)]

=2[a^2(x+y)- 1]/[(a^2x+ 1)(a^2y+ 1)]

f(x)f(y)=(a^2x- 1)/(a^2x+ 1)*(a^2y- 1)/(a^2y+ 1)

=[a^2(x+y)-a^2x-a^2y+ 1]/[(a^2x+ 1)(a^2y+ 1)]

f(x)f(y)+ 1=[a^2(x+y)-a^2x-a^2y+ 1]/[(a^2x+ 1)(a^2y+ 1)]+ 1

=[a^2(x+y)-a^2x-a^2y+ 1+(a^2x+ 1)(a^2y+ 1)]/[(a^2x+ 1)(a^2y+ 1)]

=2[a^2(x+y)+ 1]/[(a^2x+ 1)(a^2y+ 1)]

[f(x)+f(y)]/[f(x)f(y)+ 1]=[a^2(x+y)- 1]/[a^2(x+y)+ 1]=f(x+y]

So f (x+y) = [f (x)+f (y)]/[f (x) f (y)+1]