Question

A) 3^x+1 - 2 * 3^x-2 = 75 b) 8^x+1 = 5^x+1 c) 6^2x - 4 * 6^x - 12 =0 d) 9^x - 8 * 3^x = 9 найти х, заранее !

Answer 1

$3^{x+1}-2\cdot3^{x-2}=75\\\\ 3^x\cdot3-2\cdot3^x\cdot(3)^{-2}=75\\\\ 3^x(3-2\cdot(3)^{-2})=75\\\\ 3^x(3- \frac{2}{9})=75\\\\ 3^x\cdot \frac{25}{9}=75\\\\ 3^x=75\cdot \frac{9}{25} \\\\ 3^x=27\\\\ 3^x=3^3\\\\ x=3$

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$8^{x+1}=5^{x+1}\ \ \ |(\div (5^{x+1})}\\\\ \frac{8^{x+1}}{5^{x+1}}= \frac{5^{x+1}}{5^{x+1}}\\\\ ( \frac{8}{5})^{ x+1}=1\\\\ ( \frac{8}{5})^{ x+1}=( \frac{8}{5})^{0}\\\\ x+1=0\\\\ x=-1$

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$6^{2x}-4\cdot6^{x}-12=0\\\\ 6^x=t\\\\ t^2-4t-12=0\\\\ t^2-4t=12\\\\ t^2-4t+4=14+4\\\\ (t^-2)^2=16\\\\ \sqrt{(t-2)^2}=\sqrt{16}\\\\ |t-2|=4\\\\ \left \{ {{t-2=-4} \atop {t-2=4}} \right. \Longrightarrow \left \{ {{t=-2} \atop {t=6}} \right. \\\\ 6^x=-2\\ x=\varnothing\\\\ 6^x=6\\ x=1$

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$9^x-8\cdot3^x=9\\\\ 3^{2x}-8\cdot3^x=9\\\\ 3^x=t\\\\ t^2-8t=9\\\\ t^2-8t-9=0\\\\ D=64+36=100; \ \ \sqrt {D}=10\\\\ t_{1/2}= \frac{8\pm10}{2}=4\pm5\\\\ t_1=-1\\\\ t_2=9\\\\ 3^x=-1\\\\ x=\varnothing\\\\ 3^x=9\\\\ 3^x=3^2\\\\ x=2$