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umka2103 [35]
3 years ago
5

Solve 9^x-10x3^x+9=0

Mathematics
1 answer:
Goshia [24]3 years ago
4 0
Let y = 3^x , then 9^x = (3^2)^x = (3^x)^2 = y^2,  so we can write the equation as

y^2 - 10y + 9 = 0

(y - 9)(y - 1) = 0

giving y = 9 or y = 1

Therefore  3^x = 9    giving x = 2

and 3^x = 1 giving x = 0


so the solution  set is {0, 2}

Note - I have assumed that the 'x' after the 10 means 'multiply'.   Is that correct?
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Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

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b = 15.75

Step-by-step explanation:

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thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

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b = (62.5²)^{1/3} = 15.75

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3 years ago
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