JAMB 2021 Mathematics · Question 21

Simplify 2log 25\frac{2}{5} - log72125\frac{72}{125} + log 9

Options

  1. A.1 - 4 log3
  2. B.-1 + 2 log 3
  3. C.-1 + 5 log2
  4. D.1 - 2log 2

Answer: D. 1 - 2log 2

Explanation written with AI and checked against the answer key

To simplify the expression 2log⁡(25)−log⁡(72125)+log⁡(9)2\log(\frac{2}{5}) - \log(\frac{72}{125}) + \log(9), we will use the fundamental laws of logarithms.

1. Analyze the expression

The expression is: 2log⁡(25)−log⁡(72125)+log⁡(9)2\log(\frac{2}{5}) - \log(\frac{72}{125}) + \log(9)

2. Explain the concepts (Logarithm Laws)
  • Power Rule: nlog⁡(x)=log⁡(xn)n\log(x) = \log(x^n)
  • Subtraction Rule: log⁡(a)−log⁡(b)=log⁡(ab)\log(a) - \log(b) = \log(\frac{a}{b})
  • Addition Rule: log⁡(a)+log⁡(b)=log⁡(a×b)\log(a) + \log(b) = \log(a \times b)
3. Step-by-step Solution

Step 1: Apply the Power Rule to the first term 2log⁡(25)=log⁡((25)2)=log⁡(425)2\log(\frac{2}{5}) = \log((\frac{2}{5})^2) = \log(\frac{4}{25})

Now the expression is: log⁡(425)−log⁡(72125)+log⁡(9)\log(\frac{4}{25}) - \log(\frac{72}{125}) + \log(9)

Step 2: Combine the first two terms using the Subtraction Rule log⁡(425)−log⁡(72125)=log⁡(4/2572/125)\log(\frac{4}{25}) - \log(\frac{72}{125}) = \log(\frac{4/25}{72/125}) =log⁡(425×12572)= \log(\frac{4}{25} \times \frac{125}{72})

Simplify the fraction: =log⁡(11×518)=log⁡(518)= \log(\frac{1}{1} \times \frac{5}{18}) = \log(\frac{5}{18})

Step 3: Add the final term using the Addition Rule log⁡(518)+log⁡(9)=log⁡(518×9)\log(\frac{5}{18}) + \log(9) = \log(\frac{5}{18} \times 9) =log⁡(52)=log⁡(5)−log⁡(2)= \log(\frac{5}{2}) = \log(5) - \log(2)

Wait! Let's re-examine the options. Since the options involve log⁡(2)\log(2) and log⁡(3)\log(3), let's express the numbers in terms of their prime factors (2, 3, and 5).

log⁡(425)−log⁡(72125)+log⁡(9)\log(\frac{4}{25}) - \log(\frac{72}{125}) + \log(9) =(log⁡4−log⁡25)−(log⁡72−log⁡125)+log⁡9= (\log 4 - \log 25) - (\log 72 - \log 125) + \log 9 =(2log⁡2−2log⁡5)−(3log⁡2+2log⁡3−3log⁡5)+2log⁡3= (2\log 2 - 2\log 5) - (3\log 2 + 2\log 3 - 3\log 5) + 2\log 3 =2log⁡2−2log⁡5−3log⁡2−2log⁡3+3log⁡5+2log⁡3= 2\log 2 - 2\log 5 - 3\log 2 - 2\log 3 + 3\log 5 + 2\log 3

Step 4: Group like terms =(2log⁡2−3log⁡2)+(−2log⁡5+3log⁡5)+(−2log⁡3+2log⁡3)= (2\log 2 - 3\log 2) + (-2\log 5 + 3\log 5) + (-2\log 3 + 2\log 3) =−log⁡2+log⁡5= -\log 2 + \log 5

Since log⁡5=log⁡(102)=log⁡10−log⁡2=1−log⁡2\log 5 = \log(\frac{10}{2}) = \log 10 - \log 2 = 1 - \log 2: =−log⁡2+(1−log⁡2)= -\log 2 + (1 - \log 2) =1−2log⁡2= 1 - 2\log 2

4. Identify the answer

Comparing our result (1−2log⁡21 - 2\log 2) to the given options, it matches option D.

5. Exam Tip

When you see logarithms with different bases or fractions, always try to break the numbers down into their prime factors (2, 3, 5). This makes it much easier to cancel out terms and reach the simplified form required by the options!

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