Question

Как сможете 3^x+1 * 4^x=0,25 * 12^3x-1 √3^x-54 - 7 * √3^x-58 =162 6^x + 6^x+1 = 2^x + 2^x+1 + 2^x+2 9^x - 2^x+0,5 = 2^x+3,5 - 3^2x-1 5^2x-1 + 4^x = 5^2x - 4^x+1

Answer 1

1) $3^{x+1}* 4 = 0.25* 12^{3x-1}$
$12^{x}*3= \frac{1}{4}* 12^{3x-1}$
$12^{x}*3- \frac{ 12^{3x-1} }{4}=0$
$12^{x}(3- \frac{ 12^{2x-1} }{4})=0$
$3- \frac{ 12^{2x-1} }{4} =0$
$\frac{ 12^{2x-1} }{4}=3$
$12^{2x-1}=12$
2x-1 = 1
2x = 2
x=1
ответ: 1
2) $\sqrt{3} ^{x-54}-7* \sqrt{3} ^{x-58}=162$
$\sqrt{3} ^{x-54}(1-7* (\sqrt{3}) ^{-4})=162$
$\sqrt{3} ^{x-54}(1- \frac{7}{9})=162$
$\sqrt{3} ^{x-54}* \frac{2}{9}=162$
$\sqrt{3} ^{x-54}= \frac{162*9}{2}$
$3^{ \frac{x-54}{2} }= 3^{6}$
x-54 = 6*2
x-54 = 12
x = 66
ответ: 66
3) $6^{x}+ 6^{x+1}= 2^{x}+ 2^{x+1}+ 2^{x+2}$
$6^{x}(1+6) = 2^{x}(1+2+4)$
$6^{x}*7 = 2^{x}*7$
$6^{x}= 2^{x}$
$3^{x}=1$
x=0
ответ: 0
4) $9^{x}- 2^{x+0.5}= 2^{x+3.5}- 3^{2x-1}$
$3^{2x}+ 3^{2x-1}= 2^{x+3.5}+ 2^{x+0.5}$
$3^{2x}(1+ \frac{1}{3})= 2^{x}( 2^{3.5}+ 2^{0.5})$
$3^{2x}* \frac{4}{3}= 2^{x}*9 \sqrt{2}$
$3^{2x-1}* 2^{2}= 2^{x+0.5}* 3^{2}$
$3^{2x-3}= 2^{x-1.5}$
$3^{2x-3}= \sqrt{2} ^{2x-3}$
$( \frac{3}{ \sqrt{2} }) ^{2x-3} = 1$
2x-3 = 0
2x = 3
x = 1.5
ответ: 1,5
5) $5^{2x-1}+ 4^{x}= 5^{2x}- 4^{x+1}$
$5^{2x-1}- 5^{2x}= -4^{x+1}- 4^{x}$
$5^{2x}- 5^{2x-1}= 4^{x+1}+ 4^{x}$
$5^{2x}(1- \frac{1}{5})= 4^{x}(4+1)$
$5^{2x}* \frac{4}{5}= 4^{x}*5$
$5^{2x-1}*4= 4^{x}*5$
$5^{2x-2}= 4^{x-1}$
$5^{2x-2}= 2^{2x-2}$
$( \frac{5}{2}) ^{2x-2}=1$
2x-2 = 0
2x = 2
x = 1
ответ: 1