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Question: Simplify the following expression: $7-\frac{1}{2}\left[-log(10^{-5})\right]-\frac{1}{2}log(0.02)$ ...

Simplify the following expression:

7−12[−log(10−5)]−12log(0.02)7-\frac{1}{2}\left[-log(10^{-5})\right]-\frac{1}{2}log(0.02)

Answer

11−log(2)2\frac{11 - log(2)}{2}

Explanation

Solution

The given expression is 7−12[−log(10−5)]−12log(0.02)7-\frac{1}{2}\left[-log(10^{-5})\right]-\frac{1}{2}log(0.02). We assume the logarithm is base 10, i.e., log(x)=log10(x)log(x) = log_{10}(x).

We use the following properties of logarithms:

  1. log(ab)=b⋅log(a)log(a^b) = b \cdot log(a)
  2. log(a⋅b)=log(a)+log(b)log(a \cdot b) = log(a) + log(b)
  3. log(ab)=log(a)−log(b)log(\frac{a}{b}) = log(a) - log(b)
  4. log(10)=1log(10) = 1

Let's simplify the expression step by step:

7−12×(−1)×log(10−5)−12log(0.02)7 - \frac{1}{2} \times (-1) \times log(10^{-5}) - \frac{1}{2}log(0.02) =7+12log(10−5)−12log(0.02)= 7 + \frac{1}{2}log(10^{-5}) - \frac{1}{2}log(0.02)

Using property 1, log(10−5)=−5⋅log(10)log(10^{-5}) = -5 \cdot log(10). Since log(10)=1log(10) = 1, log(10−5)=−5⋅1=−5log(10^{-5}) = -5 \cdot 1 = -5.

Substitute this into the expression:

7+12(−5)−12log(0.02)7 + \frac{1}{2}(-5) - \frac{1}{2}log(0.02) =7−52−12log(0.02)= 7 - \frac{5}{2} - \frac{1}{2}log(0.02)

Now, let's simplify log(0.02)log(0.02).

0.02=2100=1500.02 = \frac{2}{100} = \frac{1}{50}. Using property 3, log(0.02)=log(150)=log(1)−log(50)log(0.02) = log(\frac{1}{50}) = log(1) - log(50). Since log(1)=0log(1) = 0, log(0.02)=0−log(50)=−log(50)log(0.02) = 0 - log(50) = -log(50).

Substitute this back into the expression:

7−52−12(−log(50))7 - \frac{5}{2} - \frac{1}{2}(-log(50)) =7−52+12log(50)= 7 - \frac{5}{2} + \frac{1}{2}log(50)

Now, let's simplify log(50)log(50).

50=5×1050 = 5 \times 10. Using property 2, log(50)=log(5×10)=log(5)+log(10)log(50) = log(5 \times 10) = log(5) + log(10). Since log(10)=1log(10) = 1, log(50)=log(5)+1log(50) = log(5) + 1.

Substitute this back into the expression:

7−52+12(log(5)+1)7 - \frac{5}{2} + \frac{1}{2}(log(5) + 1) =7−52+12log(5)+12= 7 - \frac{5}{2} + \frac{1}{2}log(5) + \frac{1}{2}

Combine the constant terms:

7−52+12=7+(1−52)=7+(−42)=7−2=57 - \frac{5}{2} + \frac{1}{2} = 7 + (\frac{1-5}{2}) = 7 + (\frac{-4}{2}) = 7 - 2 = 5.

So the expression simplifies to:

5+12log(5)5 + \frac{1}{2}log(5).

Alternatively, we can write log(5)=log(102)=log(10)−log(2)=1−log(2)log(5) = log(\frac{10}{2}) = log(10) - log(2) = 1 - log(2).

Substituting this into the expression:

5+12(1−log(2))5 + \frac{1}{2}(1 - log(2)) =5+12−12log(2)= 5 + \frac{1}{2} - \frac{1}{2}log(2) =10+12−12log(2)= \frac{10+1}{2} - \frac{1}{2}log(2) =112−12log(2)= \frac{11}{2} - \frac{1}{2}log(2) =11−log(2)2= \frac{11 - log(2)}{2}.

Therefore, the final answer is 11−log(2)2\frac{11 - log(2)}{2}.