Question

(25Б) Помогите решить 2 логорифмических уровнения

Answer 1

5) $( \log_{x} (2) )*( \log_{2x} (2) )= \log_{4x} (2)$

$x > 0, x\neq 1, x\neq \frac{1}{2} ,x\neq \frac{1}{4} .$

$[*] Formula: \log_{a} (x) =\frac{ \log_{b} (2) }{ \log_{b} (a) }.$

$\frac{ \log_{2} (2) }{ \log_{2} (x) } *\frac{ \log_{2} (2) }{ \log_{2} (2x) } =\frac{ \log_{2} (2) }{ \log_{2} (4x) }$

$\frac{1}{\log_{2} (2)\log_{2} (2x)} =\frac{1}{\log_{2} (4x)}$

$[*] Formula: \log_{a} (xy)=\log_{a} (x)+\log_{a} (y).$

$\frac{1}{\log_{2} (x)(\log_{2} (2)+\log_{2} (x))} =\frac{1}{\log_{2} (2^2)+\log_{2} (x)}$

$\log_{2} (x)*(1+\log_{2} (x))=2+\log_{2} (x)$

$\log_{2}(x)+\log_{2}(x)\log_{2}(x)=2+\log_{2}(x)$

$\log_{2}^2(x)=2 = > \left \{ {{\log_{2}(x)=\sqrt{2} } \atop {\log_{2}(x)=-\sqrt{2} }} - > \left \{ {{x_{1} =2^{-\sqrt{2} }} \atop {x_{2} =2^{\sqrt{2}}} \right.$

Ответ: $\left \{ {{x_{1} =2^{-\sqrt{2} }} \atop {x_{2} =2^{\sqrt{2}}} \right.$              все сходится.

6) $x^{\log_{a}(x)}=a^{\log_{a}^3(x)}$

$x^{\frac{ln(x)}{ln(a)} }=a^{\frac{ln^{3}(x)}{ln^{3}(a)} }$

$a > 0,x > 0,a\neq 1.$

$ln(x^{\frac{ln(x)}{ln(a)} })=ln(a^{\frac{ln^{3}(x)}{ln^{3}(a)} })$

$[*]Formula:ln(x^{y} )=y*ln(x).$

$\frac{ln^{2}(x)}{ln(a)} =\frac{ln^{3}(x)}{ln^{2}(a)}$

$\frac{ln^{2}(x)}{ln(a)} -\frac{ln^{3}(x)}{ln^{2}(a)}=0$

$\ln^{2}(x)(\frac{1}{ln(a)}-\frac{ln(x)}{\ln^{2}(a)} )=0$

$\left \{ {{\ln^{2}(x)=0} \atop {\frac{1}{ln(a)}-\frac{ln(x)}{\ln^{2}(a)} =0}$

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1) $\ln^{2}(x)=0= > e^{ln(x)}=e^{0}= > x_{1} =1.$

2) $\frac{1}{ln(a)}-\frac{ln(x)}{\ln^{2}(a)} =0$

$\frac{-ln(x)+ln(a)}{ln^{2}(a)} =0$       | *     $-ln^{2}(a)$

$ln(x)-ln(a)=0\\ln(x)=ln(a)\\e^{ln(x)}=e^{ln(a)} = > x_{2} =a.$

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Ответ: $x_{1} =1.\\x_{2} =a.$               все сходится.

Answer 2

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