Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 9

Zgjidhja e ushtrimit 9 të mësimit 2.2Z në librin Matematika 10 - 11: Pjesa I nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Zgjidhni ekuacionet e mëposhtme.

  1. (22)x×23x=32(2^2)^x \times 2^{3x} = 32
  2. (37x×35x38x)2=81(\dfrac{3^{7x} \times 3^{5x}}{3^{8x}})^2 = 81
  3. 16−x×44x83=32\dfrac{16^{-x} \times 4^{4x}}{8^3} = 32
  4. 92x×(3x)227×34x=19\dfrac{9^{2x} \times (3^x)^2}{27 \times 3^{4x}} = \dfrac{1}{9}

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Zgjidhja

  1. 22x×23x=25⇒22x+3x=25⇒25x=25⇒5x=5⇒x=12^{2x} \times 2^{3x} = 2^5 \rArr 2^{2x+3x} = 2^5 \rArr 2^{5x} = 2^5 \rArr 5x = 5 \rArr x = 1
  2. (37x×38x:38x)2=34⇒(37x+5x−8x)2=34⇒(34x)2=34⇒38x=34⇒8x=4⇒x=12\begin{aligned} (3^{7x} \times 3^{8x} : 3^{8x})^2 = 3^{4}&\rArr (3^{7x+5x-8x})^2 = 3^4 \\ &\rArr (3^{4x})^2=3^4 \\ &\rArr 3^{8x} = 3^4 \\ &\rArr 8x = 4 \\ &\rArr x = \frac{1}{2} \end{aligned}
  3. (24)−x×(22)4x(23)3=25⇒2−4x×28x29=25⇒2−4x+8x−9=25⇒24x−9=25⇒4x−9=5⇒4x=14⇒x=72\begin{aligned} \frac{(2^4)^{-x} \times (2^2)^{4x}}{(2^3)^3} = 2^5&\rArr \frac{2^{-4x} \times 2^{8x}}{2^9} = 2^5 \\ &\rArr 2^{-4x+8x-9} = 2^5 \\ &\rArr 2^{4x-9} = 2^5 \\ &\rArr 4x-9 = 5 \\ &\rArr 4x=14 \\ &\rArr x = \frac{7}{2} \end{aligned}
  4. (32)2x×32x33×34x=132⇒34x×32x34x+3=3−2⇒34x+2x−(4x+3)=3−2⇒36x−4x−3=3−2⇒32x−3=3−2⇒2x−3=−2⇒2x=1⇒x=12\begin{aligned} \frac{(3^2)^{2x} \times 3^{2x}}{3^3 \times 3^{4x}} = \frac{1}{3^2} &\rArr \frac{3^{4x} \times 3^{2x}}{3^{4x+3}} = 3^{-2} \\ &\rArr 3^{4x+2x-(4x+3)} = 3^{-2} \\ &\rArr 3^{6x-4x-3} = 3^{-2} \\ &\rArr 3^{2x-3} = 3^{-2} \\ &\rArr 2x -3 = -2 \\ &\rArr 2x = 1 \\ &\rArr x = \frac{1}{2} \end{aligned}