Показательные уравнения

Question 1

Question 2

Решите задачу:

$1)\quad 2^{12-x}=4\; \; \to \; \; \ 2^{12-x}=2^2\\\\12-x=2\; \; \Rightarrow \; \; \underline {x=10}\\\\2)\quad 8=4^{\frac{1}{12x+1}}\; \; \to \; \; 2^3=2^{\frac{2}{12x+1}}\\\\\frac{2}{12x+1}=3\; \; ,\; \; \frac{2-3(12x+1)}{12x+1}=0\; \; (x\ne -\frac{1}{12})\\\\2-36x-3=0\; ,\; \; 36x=-1\; \; ,\; \; \underline {x=-\frac{1}{36}}$

$3)\quad (\frac{12}{27} )^{\frac{x}{12}+1} =(\frac{23}{30} )^{\frac{x}{12}+1}\\\\( \frac{12}{27} \cdot \frac{30}{23} )^{\frac{x}{12}+1}=1$ $\\[tex]( \frac{40}{69} )^{\frac{x}{12}+1}=(\frac{40}{69})^0\\\\\frac{x}{12}+1=0\; \; \to \; \; \underline {x=-12}$

$4)\quad 7^{12x}-7^{12x-1}=6$

$7^{12x}-7^{12x}\cdot \frac{1}{7}=6\\\\7^{12x}\cdot (1-\frac{1}{7})=6\\\\7^{12x}\cdot \frac{6}{7}=6\\\\7^{12x}=7\\\\12x=1\\\\\underline {x=\frac{1}{12}}$

$5)\quad ( \frac{28}{38} )^{\frac{x-1}{2}}=\sqrt[12]{ \frac{38}{28} }\\\\( \frac{28}{38} )^{ \frac{x-1}{2} }=( \frac{28}{38} )^{-\frac{1}{12}}\\\\ \frac{x-1}{2} = -\frac{1}{12} \\\\12(x-1)=-2\\\\12x-12=-2\\\\12x=10\\\\\underline {x=\frac{5}{6}}$

$6)\quad 7^{x+1}-3\cdot 7^{x}=28\\\\7^{x}\cdot 7-3\cdot 7^{x}=28\\\\4\cdot 7^{x}=28\\\\7^{x}=7\\\\\underline {x=1}$

$7)\quad 2^{2x}+4\cdot 7^{x}=5\\\\4^{x}+4\cdot 7^{x}=5\\\\(\; 1+4=5\; \; \; \to \; \; pri\; x=0:\; \; 4^0+4\cdot 7^0=1+4\cdot 1=5\; )\\\\\underline {x=0}$