Question

Помогите решить, пожалуйста! 1. log(7) x - 1 = 6log(x) 7 2. log(x) 2 * log(2x) 2 = log(4x) 2 3. (0,4)^(lg^2 x - 1) = (6,25)^(-2-lg x^2)

Answer 1

1.ОДЗ:
x>0; x≠1
x∈(0;1)U(1;+∞)

$log_7x-1=6log_x7\\log_7x-1-\frac{6}{log_7x}=0\\\\log_7x=t\\t-1-\frac{6}{t}=0\ \ \ \ \ \ |*t\\t^2-t-6=0\\t_1=3;t_2=-2\\\\log_7x=3\\x=7^3\\\boxed{x=343}\\\\log_7x=-2\\x=7^{-2}\\\boxed{x=\frac{1}{49}}$



2.ОДЗ:
x>0;
x≠1;
2x≠1⇒x≠1/2
4x≠1⇒x≠1/4
x∈(0;1/4)U(1/4;1/2)U(1/2;1)U(1;+∞)

$log_x2*log_{2x}2=log_{4x}2\\\frac{1}{log_2x*log_22x}=\frac{1}{log_24x}\\log_24x=log_2x*log_22x\\log_24+log_2x=log_2x(log_22+log_2x)\\2+log_2x=log_2x(1+log_2x)\\log_2x+log_2^2x-2-log_2x=0\\log_2^2x-2=0\\log_2^2x=2\\|log_2x|=\sqrt{2}\\log_2x=\sqrt{2}\ \ \ \ \ \ \ \ \ log_2x =-\sqrt{2}\\\boxed{x=2^{\sqrt{2}}}\ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{x=2^{-\sqrt{2}}}$



3.ОДЗ:
x>0;
x∈(0;+∞)

$(0.4)^{lg^2x-1}=(6.25)^{-(lgx^2+2)}\\(0.4)^{lg^2x-1}=(2.5)^{-2(lgx^2+2)}\\(0.4)^{lg^2x-1}=((2.5)^{-1})^{2lgx^2+4}\\(0.4)^{lg^2x-1}=(0.4)^{2lgx^2+4}\\lg^2x-1=2lgx^2+4\\lg^2x-4lgx-5=0\\\\lgx=t\\t^2-4t-5=0\\t_1=5;t_2=-1\\\\lgx=5\\x=10^5\\\boxed{x=100000}\\\\lgx=-1\\x=(10)^{-1}\\\boxed{x=\frac{1}{10}}$

Answer 2

1)ОДЗ x>0 U x≠1⇒x∈(0;1) U (1;∞)log(7)x- 1=6/log(7)x
(log(7)x)²- log(7)x-6=0
log(7)x=a
a²-a-6=0
a1+a2=1 U a1*a2=-6
a1=-2⇒log(7)x=-2⇒x=1/49
a2=3⇒log(7)x=3⇒x=343
2)ОДЗ x>0 U x≠1 U x≠1/2 U x≠1/4⇒x∈(0;1/4) U (1/4;1/2) U (1/2;1)U (1;∞)
1/log(2)x *1/(log(2)x + 1)=1/(log(2) +2)
log(2)x*(log(2)x +1)=log(2)x+2
(log(2)x)²+log(2)x-log(2)x=2
(log(2)x)²=2
log(2)x=-√2⇒x=$1/2 ^{ \sqrt{2} }$
log(2)x=√2⇒x=$2 ^{ \sqrt{2} }$
3)ОДЗ x>0
$0,4 ^{lg^2x-1} =0,4 ^{2lgx^2+4}$
lg²x-1=4+4lgx
lg²x-4lgx-5=0
lgx=a
a²-4a-5=0
a1+a2=4 U a1*a2=-5
a1=-1⇒lgx=-1⇒x=0,1
a2=5⇒lgx=5⇒x=100000