SOLUTION: simplify log1 -3 log2 + log16

Question 113487: simplify log1 -3 log2 + log16
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
First recognize that log(1) = 0. [This is true regardless of what base logarithm is used.]
.
Next, work on -3*log(2). By the rules of logarithms, the multiplier of a logarithm (in this
case that multiplier is -3) can become the exponent of the quantity that the log is operating
on. That is to say -3*log(2) is equivalent to log(2^(-3)). But 2^(-3) equals 1/(2^3) by
the rules of exponents. And 1/(2^3) = 1/8 when you cube 2. Therefore, -3*log(2) converts to
log(1/8).
.
At this point the original problem has been converted to:
.
0 + log(1/8) + log(16)
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and the zero can be dropped so that the problem is just:
.
log(1/8) + log(16)
.
Recognize now that by the rules of logarithms the sum of two logarithms is equal to the
log of their product. Applying this rule to the problem results in:
.
log(1/8) + log(16) = log((1/8)*16) = log(16/8) = log(2)
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The answer to the problem is log(2)
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You can check this answer by using a scientific calculator as follows:
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a) Return to the original problem. Use the calculator to find log of 1. You should get zero
as the answer to that part of the problem.
.
b) Use the calculator to find the base 10 log of 2. The answer to that should be 0.301029995.
Multiply that by -3 and you should have -0.903089987
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c) Use the calculator to find the base 10 log of 16. You should get 1.204119983
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d) Total all the results ... 0 - 0.903089987 + 1.20411983 = 0.301029995
.
That's the result of working the problem. But we found that log(2) should be equivalent
to that. So use the calculator to find the base 10 log of 2 and you will see that it
indeed is 0.301029995 ... which verifies that log(2) is equivalent to the original problem.
.
Hope this helps you to see your way through this problem and to understand the rules of
logarithms and exponents a little better.
.

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