Question

Помогите решить, пожалуйста.

$1. 7^{x-12}=49^{x-18}$


$2. (\frac{1}{6})^{x-6}=6^{x-16}$

$13^{2-x}=(\frac{1}{13})^{5x-8}$

log4(x+21)=log8(x+21)

Answer 1

1)7^x-12 = 49^x-18

7^x-12=7^2x-36

x-12=2x-36

x=24

2)(1/6)^x-6 = 6^x-16

6^6-x=6^x-16

6-x=x-16

x=11

3) 13^2-x = 13^8-5x

2-x=8-5x

x=3/2

4)ODZ: x >-21

log4(x+21) = log8(x+21)

1/2log2(x+21)=1/3log2(x+21)

1/6log2(x+21)=0

x+21=1

x=-20

Answer 2

$1)7^{x-12}=49^{x-18}\\\\7^{x-12}=(7^{2})^{x-18}\\\\7^{x-12}=7^{2x-36}\\\\x-12=2x-36\\\\x=24$

$2)(\frac{1}{6})^{x-6}=6^{x-16}\\\\(6^{-1})^{x-6}=6^{x-16}\\\\6^{6-x}=6^{x-16}\\\\6-x=x-16\\\\2x=22\\\\x=11$

$3)13^{2-x}=(\frac{1}{13})^{5x-8}\\\\((\frac{1}{13})^{-1} )^{2-x}=(\frac{1}{13})^{5x-8}\\\\(\frac{1}{13})^{x-2}=(\frac{1}{13})^{5x-8}\\\\x-2=5x-8\\\\4x=6\\\\x=1,5$

4) ОДЗ : x + 21 > 0   ⇒   x > - 21

$log_{4}(x+21)=log_{8}(x+21)\\\\log_{4}(x+21)=\frac{2}{3} log_{4}(x+21)\\\\log_{4}(x+21)-\frac{2}{3}log_{4}(x+21)=0\\\\\frac{1}{3}log_{4}(x+21)=0\\\\log_{4}(x+21)=0\\\\x+21=1\\\\x=-20\\\\Otvet:-20$