1) log7(3x-1)=22)log2(2x-7)=log2(x+5)1) (1\2)^x=322) корень 32=4^x\2

Question

1) log7(3x-1)=2
2)log2(2x-7)=log2(x+5)

1) (1\2)^x=32
2) корень 32=4^x\2

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Answer 1

1) 3х-1=49
3х=50
Х=50/3

2)2х-7=х+5
Х=12

1) х=log[1/2](32)
X=-5

2^(5/2)=2^x
X=5/2