SOLUTION: I am having trouble solving the following problems: ln(sqrt(x-8)) = 5 log4x-log(12+(sqrt x)) = 2 The number N of trees of a given species per acre is approximated by the m

Question 314304: I am having trouble solving the following problems:
ln(sqrt(x-8)) = 5
log4x-log(12+(sqrt x)) = 2
The number N of trees of a given species per acre is approximated by the model N=68(10^-0.04x), 5 < x < 40, where x is the average diameter of the trees in inches 3 feet above the ground. Use the model to approximate the average diameter of the trees in a test plot when N=21.
Someone please help!! I cannot figure them out and would GREATLY APPRECIATE IT!!
THANK YOU!!
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
ln(sqrt(x-8)) = 5
sqrt(x-8) = e^5
(x-8) = [e^10]
x = e^10+8
x = 22034
-------------------------------
log4x-log(12+(sqrt x)) = 2
---
log[4x/[12+sqrt(x)] = 2
4x/[12+sqrt(x)] = 10^2
4x = 1200 + 100sqrt(x)
x = 300 + sqrt(x)
---
x - sqrt(x) - 300 = 0
---
Let w = sqrt(x)
Substitute to get:
w^2 - w - 300 = 0
w = [1 +- sqrt(1 - 4*-300)]/2
---
Positive solution:
w = [1 + sqrt(1201)]/2 is approximately 17.8277
---
So, sqrt(x) = 17.8277
x = 317.83
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The number N of trees of a given species per acre is approximated by the model N=68(10^-0.04x), 5 < x < 40, where x is the average diameter of the trees in inches 3 feet above the ground. Use the model to approximate the average diameter of the trees in a test plot when N=21.
----------------
N=68(10^-0.04x)
21 = 68(10^(-0.04x)
0.3088 = 10^(-0.04x)
Take the log of both sides to get:
-0.04x = log(0.3088)
x = 12.76 inches
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Cheers,
Stan H.
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