Log 5(4x+5)=2+log5(x-4). Created by shevchenkosvetl. algebra-ru

Log 5(4x+5)=2+log5(x-4)...

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$log_5(4x+5)=2+log_5(x-4) \\ ODZ: \left \{ {{x>4} \atop {x>- \frac{5}{4} }} \right. \\ log_5(4x+5)=log_55^2+log_5(x-4) \\ 4x+5=25(x-4) \\ 21x=105 \\ x=5- otvet$

3 votes Thanks 6

gusev1948

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Log5(4x+5)-Log5(x-4)=2
Log5(4x+5)/(x-4)=2 - разность логарифмов равна логарифму частного
(4x+5)/(x-4)=25 (25 - основание 5 во второй степени)
4x+5=25x-100 105=21 х=5
для проверки: Log5(20+5)=2
Log5(5-4)=Log5(1)=0 5 в нулевой степени ровно единице

2 votes Thanks 5

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