Question 1193234: given log 7=0.8451 and log 2 =0.3010, find the value of log 28.
Found 4 solutions by Alan3354, Theo, ikleyn, MathTherapy:
Answer by Alan3354(69443)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you are given:
log(7) = .8451
log(2) = .3010
you want to find the log of (28).
since 28 = 7 * 2 * 2, then you want to find the log of (7 * 2 * 2)
by the log rules, log(7 * 2 * 2) = log(7) + log(2) + log(2) = .8451 + .3010 + .3010 = 1.4471.
your solution is that log(28) = 1.4471.
to confirm, i dig the following:
the values given to you are rounded to 4 decimal places.
use your calculator find the more detailed numbers.
log(7) = .84509804...
log(2) = .3010299957...
you get log(7) + log(2) + log(2) = 1.447158031, using the values for log(7) and log(2) stored in the calculator.
using you calculator, you get log(28) = 1.447158031.
that's the same as log(7) + log(2) + log(2), confirming the procedure is correct.
the procedure was:
log(a * b * b) = log(a) + log(b) + log(b)
when a is 7 and b is 2, this becomes log(7) + log(2) + log(2) is equal to log(7 * 2 * 2) which is equal to log(28).
bottom line:
log(28) = log(7 * 2 * 2) = log(7) + log(2) + log(2) = .8451 + .3010 + .3010 = 1.4471.
here is a reference on the rules of logarithm arithmetic.
https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/
Answer by ikleyn(52903)  (Show Source):
Answer by MathTherapy(10557)  (Show Source):
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