Libro De Baldos

1.- hallar el valor de 1215 x 0.84 por algoritmos
Log (1215 x 0.84 (=log 1215 + log 0.84
=3.0844576 + 1.92479
=3.008855
1215 x 0.84= 1020.59
1020.6
2.-hallar por log elvalor de 3214.8 x 0.003 x (-43.76)
Log (3214.8 x 0.003 x43.76) = log 3214.8 + log 0.003 + log 43.76
=3.507154 + 3.477121 + 1.641077
=2.625352
3214.8 x 0.003 x (-43.76) =-422.0388
3.-hallar el valor de 0.765/39.14 por log.
Log 0.765/39.14 = log 0.76 – log 39.14
Log 0.765/39.14= Log 0.765 + colog 39.14
=1.883661 + 2.407379
=2.291040
2.291040 0.765/39.14 =0.019545
4.-dados log 2 = 0.301030 y log 3 = 0.177121 hallar log 108 sin usar la tabla
108 = 22 x 33
Log 108 = 2 (log 2) + 3 (log 3)
=2 (0.301030) + 3(0.477121)
=0.602060 + 1.4313663=2.033423
5.-dado log 115 = 2.33423 y log 5 = 0.698970 hallar log 23
23 = 115/5
Log 23 = log 115 + colog 5
= 2.060698 + 1.301030
=1.361728
6.-resolver la ecuación 3x =60
X(log 3) = log 60
X = log 60/ log 3 = 1.778151/ 0.477121 = 3.72
7.-resolvar la ecuación 52x-1 =125
(2 x -1) log 5 = log 125
2x-1 = log 125 / log 5
2x = log 125 / log 5 +1
Log 125 / log 5 +1X = log 5 / 2
X = 2.096910 / 0.698970 / 2 + 1
= 3 + 1 /2 = 2
8.-hayar el cociente de x15 – y 15 entre x3 – y 3
x15 – y 15 / x3 – y 3 =x12 + x 9 y3 + x6 + y6 + x3 y9 + y12
9.-resolverla ecuación 3x – 2x/5 = x/10 – 7/4
60x – 8x = 2x – 35
60x – 8x - 2x = - 35 x = - 35/50 = - 7/10
50x = -35
10.-resolver 3/2x+1 – 2/2x-1 – x+3/4x2-1 = 0
3(2x-1) – 2 (2x+1) – (x+3) = 0
6x - 3- 4x - x - 3 = 0
6x - 4x – x = 3 + 2 + 3
X = 8
11.- hallar el área de un trapecio cuya altura mide 5 cm y sus bases 6 y 8 m.
A = h (b + b´ / 2)
H = 5 m B = 6 m B´ = 8 m
A =5 (6+8/2) = 5 x7 = 35 m2

12.-hallar el volumen de una pirámide siendo su altura 12 m y el area de la base 36 m2
V = 1/1 h x b
H = 12 m, b = 36 m2
V = 1/3 x 12 x 36 = 4 x 36 = 144 m3
13.-ne la formula 1/f...