Please help me answer this..
Suppose that
2^(x/2)= p
3^(x/3-1)= q
[5^(x/2)]/3 = r
Express (0.48)^x in terms of p, q and/ or r.
this question came from our reviewer.. please and thank you!
.48 = 48/100 = 12/25 = 2*2*3/(5*5)
if 2^(x/2)= p then 2 = p^(2/x)
if 3^(x/3-1)= q or 3^(x-3)/3) then 3 = q^(3/(x-3))
if [5^(x/2)]/3 = r then 5 = (3r)^(2/x)
so .48 = 2*2*3/(5*5)
= (p^(2/x))(p^(2/x)(q^(3/(x-3)))/[(3r)^(2/x)*(3r)^(2/x)]
= 2^x * 3^(3/(x-3)) / (3r)^x
forgot to finish it, since it was .48^x
so my answer 2^x * 3^(3/(x-3)) / (3r)^x raised to the x
or [2^x * 3^(3/(x-3)) / (3r)^x]^x
To express (0.48)^𝑥 in terms of 𝑝, 𝑞, and/or 𝑟, we need to manipulate and simplify the given equations.
Let's start by expressing each equation in terms of 𝑥.
Equation 1: 2^(𝑥/2) = 𝑝
To solve for 𝑥, we can take the logarithm of both sides with base 2:
𝑥/2 = log2(𝑝)
𝑥 = 2 * log2(𝑝)
Equation 2: 3^(𝑥/3-1) = 𝑞
Similarly, we take the logarithm with base 3:
𝑥/3 - 1 = log3(𝑞)
𝑥/3 = log3(𝑞) + 1
𝑥 = 3 * (log3(𝑞) + 1)
Equation 3: (5^(𝑥/2))/3 = 𝑟
First, let's remove the fraction by multiplying both sides by 3:
5^(𝑥/2) = 3𝑟
Now, we can take the logarithm with base 5:
𝑥/2 = log5(3𝑟)
𝑥 = 2 * log5(3𝑟)
Now that we have expressions for 𝑥 in terms of 𝑝, 𝑞, and 𝑟, let's substitute them into (0.48)^𝑥:
(0.48)^𝑥 = (0.48)^(2 * log2(𝑝)) [using the expression for 𝑥 from Equation 1]
= [(0.48)^2]^(log2(𝑝))
(0.48)^𝑥 = (0.48)^(3 * (log3(𝑞) + 1)) [using the expression for 𝑥 from Equation 2]
= [(0.48)^3]^(log3(𝑞) + 1)
(0.48)^𝑥 = (0.48)^(2 * log5(3𝑟)) [using the expression for 𝑥 from Equation 3]
= [(0.48)^2]^(log5(3𝑟))
Therefore, we can express (0.48)^𝑥 in terms of 𝑝, 𝑞, and 𝑟 as:
(0.48)^𝑥 = [(0.48)^2]^(log2(𝑝)) = [(0.48)^3]^(log3(𝑞) + 1) = [(0.48)^2]^(log5(3𝑟))