Real Numbers (SAQs)

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SAQ-1 : Express 140 and 3825 as product its prime factors

$$140 = 2 \times 2 \times 7 \times 5$$

$$= 2^2 \times 7 \times 5$$

$$3825 = 3 \times 1275$$

$$= 3 \times 3 \times 425$$

$$= 3 \times 3 \times 5 \times 85$$

$$= 3^2 \times 5 \times 5 \times 17$$

$$= 3^2 \times 5^2 \times 17$$

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SAQ-2 : Expand log 343/125

$$\log \frac{343}{125} = \log 343 – \log 125$$

$$= \log 7^3 – \log 5^3$$

$$= 3 \log 7 – 3 \log 5$$

$$\log \frac{343}{125} = 3(\log 7 – \log 5)$$

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SAQ-3 : Solve 3^x = 5^x-2

$$3^x = 5^{x-2}$$

$$\log 3^x = \log 5^{x-2}$$

$$x \log 3 = (x – 2) \log 5$$

$$x \log 3 = x \log 5 – 2 \log 5$$

$$x \log 5 – x \log 3 = 2 \log 5$$

$$x(\log 5 – \log 3) = 2 \log 5$$

$$x = \frac{2 \log 5}{\log 5 – \log 3}$$

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SAQ-4 : Check whether 6ⁿ can end with the digit 0 for any natural number n

$$\text{For } 6n \text{ to end with the digit ‘0’, it must have ‘5’ as a factor.}$$

$$\text{But we know that the prime factors of } 6n \text{ are } 2 \text{ and } 3 \text{ only.}$$

$$\text{Therefore, } 6n \text{ can never end with the digit ‘0’.}$$

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SAQ-5 : If x² + y² = 10xy prove that 2 log(x + y) = log x + log y + 2 log 2 + log 3

$$x^2 + y^2 = 10xy$$

$$x^2 + y^2 + 2xy = 10xy + 2xy$$

$$(x + y)^2 = 12xy$$

$$\log (x + y)^2 = \log 12xy$$

$$2 \log (x + y) = \log 12xy$$

$$= \log 12 + \log x + \log y$$

$$= \log (4 \times 3) + \log x + \log y$$

$$= \log 4 + \log 3 + \log x + \log y$$

$$= \log 2^2 + \log 3 + \log x + \log y$$

$$= 2 \log 2 + \log 3 + \log x + \log y$$

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SAQ-6 : Lalitha says that HCF and LCM of the numbers 80 and 60 are 20 and 120 respectively. Do you agree with her? Justify

$$\text{Given two numbers are } 80, 60$$

$$\text{HCF = 20 and LCM = 120}$$

$$\text{HCF} \times \text{LCM} = \text{product of two numbers}$$

$$80 \times 60 = 4800$$

$$20 \times 120 = 2400$$

$$4800 \neq 2400$$

$$\text{These two numbers are not equal. So, I did not agree with Lalitha.}$$

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SAQ-7 : Show that log 162/343 + 2 log 7/9 – log 1/7 = log 2

$$\log \frac{162}{343} + 2 \log \frac{7}{9} – \log \frac{1}{7}$$

$$= \log 162 – \log 343 + 2(\log 7 – \log 9) – (\log 1 – \log 7)$$

$$= \log (81 \times 2) – \log 7^3 + 2 \log 7 – 2 \log 3^2 – \log 1 + \log 7$$

$$= \log 81 + \log 2 – 3 \log 7 + 2 \log 7 – 4 \log 3 – \log 1 + \log 7$$

$$= \log 3^4 + \log 2 – 4 \log 3 – \log 1$$

$$= 4 \log 3 + \log 2 – 4 \log 3 – \log 1$$

$$= \log 2 – \log 1$$

$$= \log 2$$