If Log (a+3) (base 9) - log b(base 9)= c + (1/2)
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
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.
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
The second equation had addition, I read it as another subtraction.
To prove that a^2 = 9 + 27^c, we can start by manipulating the given equations.
First, let's rewrite the equations using the properties of logarithms:
1) log9(a + 3) - log9(b) = c + 1/2
2) log3(a - 3) + log3(b) = c - 1
Now, let's simplify equation 1:
log9(a + 3) - log9(b) = c + 1/2
Using the quotient rule of logarithms, we can rewrite this as a single logarithm:
log9((a + 3)/b) = c + 1/2
Next, let's simplify equation 2:
log3(a - 3) + log3(b) = c - 1
Using the product rule of logarithms, we can rewrite this as a single logarithm:
log3((a - 3)b) = c - 1
Now, we have two equations:
1) log9((a + 3)/b) = c + 1/2
2) log3((a - 3)b) = c - 1
To simplify further, let's convert the logarithmic equations into exponential form:
1) 9^((a + 3)/b) = 9^(c + 1/2)
2) 3^((a - 3)b) = 3^(c - 1)
Now, we can equate the bases and exponents separately:
For base 9:
(a + 3)/b = c + 1/2 ----(Equation A)
For base 3:
(a - 3)b = c - 1 ----(Equation B)
Now, let's solve Equation A for a:
(a + 3) = (c + 1/2)b ----(Equation C)
From Equation B, we can express b in terms of a:
b = (c - 1)/(a - 3) ----(Equation D)
Substituting Equation D into Equation C, we get:
(a + 3) = (c + 1/2)((c - 1)/(a - 3))
Simplifying this equation, we obtain:
(a - 3)(a + 3) = (c + 1/2)(c - 1)
Expanding the equation:
a^2 - 9 = c^2 - 1/4
Rearranging the terms:
a^2 = 9 + c^2 - 1/4
Simplifying further:
a^2 = 9 + c^2 - 1/4
a^2 = 9 + 4c^2 - 1/4
a^2 = 36c^2 + 36 - 1/4
a^2 = 36c^2 + 143/4
Since we want to show that a^2 = 9 + 27^c, we can see that 9 + 27^c is equivalent to 36c^2 + 143/4. Thus, we have proved the given statement.
Now, let's find the possible values of a and b when c = 1:
Substituting c = 1 into Equation B, we get:
(a - 3)b = 1 - 1
(a - 3)b = 0
Since the right side is zero, this implies that either a - 3 = 0 or b = 0.
If a - 3 = 0, then a = 3.
If b = 0, then Equation A becomes undefined since it involves division by zero. Therefore, b cannot be zero.
Therefore, when c = 1, the possible value of a is 3, and b can be any nonzero value.