Помогите решить:* 1.Lgx=2+lg3-lg5 2.log2X+log8X=8 3.log3(1-2x)=1. Created by Maria0805. algebra-ru

Помогите решить:* 1.Lgx=2+lg3-lg5 2.log2X+log8X=8 3.log3(1-2x)=1...

1). $lg x=2+lg_3-lg_5;\\ x>0;\\ lg x=lg 10^2+lg_3-lg_5;\\ lg x=lg 100+lg_3-lg_5;\\ lg x=lg \frac{100*3}{5};\\ lg x=lg 60;\\ x=60$

2. $log_2 x+log_8 x=8;\\ x>0;\\ log_2 x+log_{2^3} x=8;\\ log_2 x+\frrac{1}{3} log_2 x=8;\\ 3log_2 x+log_2 x=24;\\ 4log_2 x=24;\\ log_2 x=24:4;\\ log_2 x=6;\\ x=2^6;\\ x=64$

3. $log_3 (1-2x)=1;\\ 1-2x>0;1>2x;0.5>x;\\ x$

dtnth

Verified answer

1.Lgx=2+lg3-lg5

Lgx=2lg10+lg3/5

Lgx=lg100*3/5

Lgx=lg60

x=60

2.log2X+log8X=8

log2X+1/3*log2X=8

log2X*X^1/3=8

x^4/3=2^8

x=2^6

x=64

3.log3(1-2x)=1

1-2x=3

2x=-2

x=-1

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