Question

Решить 4 примера по теме "Степени и логарифмы". Буду очень благодарна за помощь.

Answer 1

5.58
3*2^x-2*2^x/2 -1=0
2^x/2=a
3a²-2a-1=0
D=4+12=16
a1=(2-4)/6=-1/3⇒2^x/2=-1/3-нет решения
a2=(2+4)/6=1⇒2^x/2=1⇒x/2=0⇒x=0
5.60
3^2x-2<5*3^x*2^x+2^2x=0 /2^2x≠0
(3/2)^2x-2,5*(3/2)^x+1=0
(3/2)^x=a
a²-2,5a+1=0
D=6,25-4=2,25
a1=(2,5-1,5)/2=1/2⇒(3/2)^x=1/2⇒x=log(1,5)0,5
a2=(2,5+1,5)/2=2⇒(3/2)^x=2⇒x=log(1,5)2
5.62
4*2^2x-2^x*3^x-18*3^2x=0 /3^2x≠0
4*(2/3)^2x-(2/3)^x-18=0(2/3)^x=a
4a²-a-18=0
D=1+288=289
a1=(1-17)/8=-2⇒(2/3)^x=-2-нет решения
a2=(1+17)/8=9/4⇒(2/3)^x=9/4⇒x=-2
5,64
2^12(x²+18)=2^28(x²+2x)
12(x²+18)=28(x²+2x)
3(x²+18)=7(x²+2x)
3x²+54=7x²+14x
7x²+14x-3x²-54=0
4x²+14x-54=0
2x²+7x-27=0
D=49+216=265
x1=(-7-√265)/4
x2=(-7+√265)/4

Answer 2

$3*2^x-2^{\frac{x}{2}+1}=1$
$2^{\frac{{x}{2}}=t>0;2^x=t^2$
$3t^2-2t-1=0$
$(t-1)(3t+1)=0$
$t-1=0;t_1=1$
$3t+1=0;t_2=-\frac{1}{3}<0$
$t=1$
$2^{\frac{x}{2}}=1$
$2^{\frac{x}{2}}=2^0$
$\frac{x}{2}=0$
$x=0*2$
$x=0$
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$9^x+4^x=2.5*6^x$  |:4^x
$((\frac{3}{2})^x)^2-2.5*(\frac{3}{2})^x+1=0$
$1.5^x=t>0$
$t^2-2.5t+1=0$
$2t^2-5t+2=0$
$(2t-1)(t-2)=0$
$2t-1=0;t=0.5$
$t-2=0;t_2=2$
$1.5^x=0.5;x_1=log_{1.5} 0.5$
$1.5^x=2; x_2=log_{1.5} 2$
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$4^{x+1}-6^x-2*9^{x+1}=0;$ |:4^x
$-2*9*((\frac{3}{2})^x)^2-(\frac{3}{2})^x+4*1=0$
$1.5^x=t>0;$
$-18t^2-t+4=0$
$18t^2+t-4=0$
$D=1^2-4*18*(-4)=289=17^2$
$t_1=\frac{-1-17}{2*18}<0$
$t_2=\frac{-1+17}{2*18}=\frac{4}{9}=(\frac{3}{2})^{-2}$
$(\frac{3}{2})^x=(\frac{3}{2})^{-2}$
$x=-2$
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$8^{4(x^2+8)}=16^{7(x^2+2x)}$
$2^{3*4(x^2+8)}=2^{4*7(x^2+2x)}$
$3*4(x^2+8)=4*7(x^2+2x)$
$3(x^2+8)=7(x^2+2x)$
$3x^2+24=7x^2+14x$
$4x^2+14x-24=0$
$2x^2+7x-12=0$
$D=7^2-4*2*(-12)=145$
$x_{1,2}=\frac{-7^+_-\sqrt{145}}{4}$