Question

Решите уравнение
$4 ^{3x-1} -2* 5^{2-3x} -7+ \frac{6* 20^{2-3x}+8*5 ^{2-3x}+16 }{3*4 ^{2-3x}+4 } =0$

Answer 1

$t = 2-3x; \ \ \ \ \ 3x =2-t \\ \\ 4^{2-t-1} - 2\cdot 5^t - 7 + \frac{6 \cdot 20^t + 8 \cdot 5^t +16}{3 \cdot 4^t +4}=0 \\ \\ 4^{1-t} + \frac{-2\cdot 5^t \cdot (3 \cdot 4^t +4) + 6 \cdot 20^t + 8 \cdot 5^t +16}{3 \cdot 4^t +4}-7=0 \\ \\ 4^{1-t} + \frac{-6\cdot 20^t -8\cdot 5^t + 6 \cdot 20^t + 8 \cdot 5^t +16}{3 \cdot 4^t +4}-7=0 \\ \\ \frac{4}{4^t} + \frac{16}{3 \cdot 4^t +4}-7=0 \ \ \ \cdot | \ 4^t \cdot (3 \cdot 4^t +4)\\ \\ 12 \cdot 4^t +16 +16 \cdot 4^t - 7 \cdot 4^t \cdot (3 \cdot 4^t +4)=0$

$\\ \\ 28 \cdot 4^t - 21 \cdot 4^{2t} -28 \cdot 4^t +16 =0 \\ \\ -21 \cdot 4^{2t} =-16 \\ \\ 21 \cdot 4^{2t}=16; \ \ \ \ 4^{2t}=\frac{16}{21}; \ \ \ 2t=\log_4 \frac{16}{21}; \ \ \ t=\frac{1}{2} \cdot \log_{2^2}\frac{16}{21}; \\ \\ t=\frac{1}{4} \cdot \log_2 16 - \frac{1}{4} \log_2 21=1 -\frac{\log_2 21}{4} \\ \\ 2-3x = 1 - \frac{\log_2 21}{4}; \ \ \ 3x=1 + \frac{\log_2 21}{4}=\frac{\log_216 + \log_2 21}{4}=\frac{\log_2336}{4} \\ \\ x =\frac{\log_2 336}{12}$

Answer 2

$4^{3x-1} -2* 5^{2-3x} -7+ \frac{6* 20^{2-3x} +8* 5^{2-3x}+16 }{3* 4^{2-3x}+4 } =0$

$\frac{ 4^{3x-1} *({3* 4^{2-3x}+4})}{{3* 4^{2-3x}+4}} - \frac{2* 5^{2-3x({3* 4^{2-3x}+4})} }{{3* 4^{2-3x}+4}} - \frac{7*({3* 4^{2-3x}+4})}{{3* 4^{2-3x}+4}} +$
$+\frac{6* 20^{2-3x} +8* 5^{2-3x}+16 }{3* 4^{2-3x}+4 } =0$

ОДЗ:
$3* 4^{2-3x} +4 \neq 0$

$3* 4^{3x-1+2-3x} + 4^{3x-1+1} -6* 20^{2-3x}-8* 5^{2-3x} -21* 4^{2-3x} -28+$$6* 20^{2-3x} +8* 5^{2-3x} +16=0$

$3*4+ 4^{3x} -28+16-21* 4^{2-3x} =0$

$12+ 4^{3x} -28+16-21* 4^{2-3x} =0$

$4^{3x} -21* 4^{2-3x} =0$

$4^{3x} -21* \frac{4^2}{ 4^{3x} } =0$

замена : $4^{3x} =t$ ,  $t\ \textgreater \ 0$

$t- \frac{336}{t}=0$
$t \neq 0$

$t^2-336=0$

$(t- \sqrt{336} )(t+ \sqrt{336} )=0$

$t= \sqrt{336}$  или $t=- \sqrt{336}$ не подходит

$4^{3x} = \sqrt{336}$

$3x= log_{4} \sqrt{336}$

$3x= log_{2^2} \sqrt{336}$

$3x= \frac{1}{2} log_{2} \sqrt{336}$

$3x= \frac{1}{2} log_{2} 336^{ \frac{1}{2} }$

$3x= \frac{1}{4} log_{2} 336$

$3x= \frac{log_{2} 336}{4}$

$x= \frac{log_{2} 336}{12}$

Ответ: $x= \frac{log_{2} 336}{12}$