Ejercicio 301

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CAPITULO XXXVIII

Logaritmos

Resolver las ecuaciones:

5 x =3 log 5 x =log3 xlog5 =0.4771212547 x( 0.6989700043 ) =0.4771212547 x = 0.4771212547 0.6989700043 x =0.6826061945
7 x =512 log 7 x =log512 xlog7 =2.709269961 x( 0.84509805 ) =2.709269961 x = 2.709269961 0.84509805 x =3.205864684
0. 2 x =0.0016 log0. 2 x =log0.0016 xlog0.2 =–2.795880017 x( –0.6989700043 ) =–2.795880017 x = –2.795880017 –0.6989700043 x =4
9 x =0.576 log 9 x =log0.576 xlog9 =–0.2395775166 x( 0.9542425094 ) =–0.2395775166 x = –0.2395775166 0.9542425094 x =–0.2510656507
3 x+1 =729 log 3 x+1 =log729 ( x+1 ) log3 =2.862727528 ( x+1 ) ( 0.4771212547 ) =2.862727528 x+1 = 2.862727528 0.4771212547 x =6–1 x =5
5 x–2 =625 log 5 x–2 =log625 ( x–2 ) log5 =2.795880017 ( x–2 ) ( 0.6989700043 ) =2.795880017 x–2 = 2.795880017 0.6989700043 x =4+2 x =6
2 3x+1 =128 log 2 3x+1 =log128 ( 3x+1 ) log2 =2.10720997 ( 3x+1 ) ( 0.3010299957 ) =2.10720997 3x+1 = 2.10720997 0.3010299957 3x =7–1 x = 3 x =2
3 2x–1 =2187 log 3 2x–1 =log2187 ( 2x–1 ) log3 =3.339848783 ( 2x–1 ) ( 0.4771212547 ) =3.339848783 2x–1 = 3.339848783 0.4771212547 2x =7+1 x = 2
1 1 2x =915 log1 1 2x =log915 ( 2x ) log11 =2.961421094 2x( 1.041392685 ) =2.961421094 2x = 2.961421094 1.041392685 2x =2.84371221 x = 2.84371221 2 x =1.421856105