If Log (a+3) (base 9) - log b(base 9)= c + (1/2)

Question

and log (a-3) (base 3) + log b (base 3)=c - 1.
show that a^2=9 + 27^c and find the possible
values of a and b when c=1

Answer 1

log_9(a+3)-log_9(b) = c + (1/2)
log_3(a-3)+log_3(b) =c - 1

(a+3)/b = 9^(c+1/2)
(a-3)b = 3^(c-1)

(a+3)/b = 3*9^c
(a-3)b = 3^c/3

a+3 = 3b*3^(2c)
3(a-3)b = 3^c

3(a+3)(a-3)b = 3b*3^(3c)
a^2-9 = 27^c
a^2 = 9+27^c

I expect you can get the rest.

Answer 2

Log (a+3) (base 9) - log b(base 9)= c + (1/2)
log₉ (a+3) - log₉ b = c+1/2
log₉ ((a+3)/b) = c+1/2
---> 9^(c+1/2) = (a+3)/b
3^(2c+1) = (a+3)/b **

log (a-3) (base 3) + log b (base 3)=c - 1
log₃ (a-3)+ log₃ b = c-1
log₃ ( (a-3)/b )= c-1
--> 3^(c-1) = (a-3)/b ***

divide ** by ***
3^(c+2) = (a+3)/(a-3)
(3^c)(3^2) = (a+3)/(a-3)
9(3^c) = (a+3)/(a-3)

if c = 1
27 = (a+3)/(a-3)
27a - 81 = a+3
26a = 84
a = 42/13
in **
3^(2c+1) = (a+3)/b
27 = (42/13 + 3)/b
27b = 81/13
b = 3/13

I was expecting a nicer answer.
I did not write this out on paper first, so I might have made an error
Please check my arithmetic

Answer 3

The second equation had addition, I read it as another subtraction.