Logaritmos resueltos (4º eso)

Problemas de Matem´ticas 4o de ESO a
Logaritmos Resuelve las siguientes expresiones logar´ ıtmicas: 1. log(2x) = −1 Soluci´n: o 2x = 10−1 =⇒ 1 2. log x2 = 2 Soluci´n: o x2 = 10 2 =⇒ x =
1

1 2010 2 = (10 2 ) 2 = 10 4 √ 4 x = 10

1

1

1

1

3. log(3x) = 2 Soluci´n: o 3x = 102 =⇒ x = 4. log x = −1 Soluci´n: o x = 10−1 = 3 5. log x = 2 Soluci´n: o 6. log(3x2 ) = −2 Soluci´n: o100 3

1 10

x = 10 2 =

3

√

√ 103 = 10 10

x=

1 300 √ √ 3 1 1 3 1 √ = = √ = =√ 2 30 300 10( 3) 300 10 3 3x2 = 10−2 =⇒ x2 =

7. log x + log 50 = log 100 Soluci´n: o log(x · 50) =log 1000 50x = 1000 =⇒ x = 20 8. log x = 1 + log(22 − x) Soluci´n: o log x = log 10 + log(22 − x) =⇒ log x = log 10(22 − x) x = 10(22 − x) =⇒ x = 220 − 10x =⇒ 11x = 220 =⇒ x = 220 = 20 11

1

9. 2log x − log(x − 16) = 2 Soluci´n: o log x2 − log(x − 16) = log 100 =⇒ log x2 = log 100 x − 16

x2 = 100 =⇒ x2 = 100(x − 16) =⇒ x2 = 100x − 1600 =⇒ x − 16 x2 − 100x + 1600 = 0 =⇒ x = 80, x = 20 10.log x3 = log 6 + 2 log x Soluci´n: o log x3 = log 6 + log x2 =⇒ log x3 = log(6x2 ) x3 = 6x2 =⇒ x3 − 6x2 = 0 =⇒ x2 (x − 6) = 0 =⇒ x = 0, x = 6 La soluci´n x = 0 no es v´lida. o a 11. 3 log x + 2 log x2= log 128 Soluci´n: o log x3 + log x4 = log 128 =⇒ log x7 = log 128 x7 = 128 =⇒ x7 = 27 =⇒ x = 2 12. 5 log(2x) = 20 Soluci´n: o log(2x) = 4 =⇒ 2x = 104 = 10000 =⇒ x = 5000 13. log 2x − 4 5 Soluci´n: o=2 log 2x − 4 5 = log 100

2x − 4 = 100 =⇒ 2x − 4 = 500 =⇒ 2x = 504 =⇒ x = 252 5 14. log(7x + 15) − log 5 = 1 Soluci´n: o log

7x + 15 = log 10 5

7x + 15 = 10 =⇒ 7x + 15 = 50 =⇒ 7x = 35 =⇒ x =5 5

2

x 15. log = 1 + log(21 − x) 2 Soluci´n: o x x log = log 10 + log(21 − x) =⇒ log = log 10(21 − x) 2 2 x = 10(21 − x) =⇒ x = 20(21 − x) =⇒ x = 420 − 20x =⇒ 2 21x = 420 =⇒ x = 20 10 16. log= 2 − 2 log x x Soluci´n: o log 10 10 100 = log 100 − log x2 =⇒ log = log 2 x x x

10 100 = 2 =⇒ 10x2 = 100x =⇒ 10x2 − 100x = 0 x x =⇒ 10x(x − 10) = 0 =⇒ x = 0, x = 10 La soluci´n x = 0 no es...