To find all ordered pairs (a, b) such that:
1. a^29^b = 4
2. a / 3^b = 18
We can start by solving the second equation for 'a' in terms of 'b' and then substitute it into the first equation.
From the second equation, we have:
a = 18 * 3^b
Now, substitute this expression for 'a' into the first equation:
(18 * 3^b)^(29^b) = 4
Now, let's simplify the left side:
(18 * 3^b)^(29^b) = 4
Now, raise both sides to the power of (1/(29^b)) to isolate the expression in parentheses:
(18 * 3^b) = 4^(1/(29^b))
Now, simplify the right side:
(18 * 3^b) = 4^(1/(29^b))
4^(1/(29^b)) is equivalent to taking the 29^b-th root of 4. Let's simplify that:
(18 * 3^b) = 4^(1/(29^b)) = (4^(1/29))^b
Now, we have:
18 * 3^b = (4^(1/29))^b
Now, we can set the bases equal to each other:
18 * 3^b = 4^(1/29)
To find the value of 'b,' we can first solve for '3^b':
3^b = (4^(1/29)) / 18
Now, take the natural logarithm (ln) of both sides:
ln(3^b) = ln((4^(1/29)) / 18)
Use the properties of logarithms to simplify:
b * ln(3) = ln(4^(1/29)) - ln(18)
Now, isolate 'b':
b = (ln(4^(1/29)) - ln(18)) / ln(3)
Now, we can calculate the value of 'b' using the natural logarithm and then find 'a' using the second equation:
b ≈ -3.472
Now, use this value of 'b' in the second equation to find 'a':
a = 18 * 3^b
a ≈ 18 * 3^(-3.472)
a ≈ 0.017
So, the ordered pair (a, b) that satisfies both equations is approximately (0.017, -3.472).