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Find all ordered pairs (a,b) such that a^29^b=4 and a/3^b=18

User Maumercado
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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).

User Nealium
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