Real Numbers (VSAQs)
VSAQ-1 : Find the value of log5125
$$\text{Let } x = \log_5 125$$
$$= \log_5 5^3$$
$$= 3 \log_5 5$$
$$= 3$$
VSAQ-2 : If HCF (a, b) = 12 and a x b = 1800, then find LCM (a, b)
$$\text{Given, HCF }(a, b) = 12$$
$$a \times b = 1800$$
$$\text{We know that LCM }(a, b) \times \text{HCF }(a, b) = a \times b$$
$$\text{LCM }(a, b) = \frac{a \times b}{\text{HCF }(a,b)} = \frac{1800}{12}$$
$$= 150$$
VSAQ-3 : Find H.C.F and L.C.M of 220 and 284 by prime factorisation method
$$220 = 2 \times 2 \times 5 \times 11$$
$$= 2^2 \times 5 \times 11$$
$$284 = 2 \times 2 \times 71$$
$$= 2^2 \times 71$$
$$\text{HCF }(220, 284) = 2^2 = 4$$
$$\text{LCM }(220, 284) = 2^2 \times 5 \times 11 \times 71 = 15620$$
VSAQ-4 : Expand log 128/625
$$\log \frac{128}{625} = \log 128 – \log 625$$
$$= \log 2^7 – \log 5^4$$
$$= 7 \log 2 – 4 \log 5$$
VSAQ-5 : Expand log a3b2c5
$$\log(a^3 b^2 c^5) = \log(a^3) + \log(b^2) + \log(c^5)$$
$$= 3\log(a) + 2\log(b) + 5\log(c)$$
VSAQ-6 : Find the HCF of 24 and 33 by using division algorithm
$$\text{Since } 33 > 24$$
$$\text{We apply Euclid division algorithm to } 33 \text{ and } 24$$
$$33 = 24 \times 1 + 9$$
$$24 = 9 \times 2 + 6$$
$$9 = 6 \times 1 + 3$$
$$6 = 3 \times 2 + 0$$
$$\text{HCF of } 24 \text{ and } 33 \text{ is } 3$$
VSAQ-7 : Determine the value of log100.01
$$\log_{10} 100.01 = \log_{10} \left(\frac{1}{100}\right)$$
$$= \log_{10} \left(\frac{1}{10^2}\right)$$
$$= \log_{10} \left(10^{-2}\right)$$
$$= -2 \log_{10} 10$$
$$= -2$$
VSAQ-8 : Ramu says, if log10x = 0, value of x = 0. Do you agree with him? Give reason
$$\text{Given, } \log_{10} x = 10$$
$$x = 10^{10}$$
$$x = 0$$
$$\text{If } \log_{10} x = 0 \text{ then,}$$
$$10^0 = x$$
$$x = 1 \neq 0$$
$$\text{So, I don’t agree with Ramu.}$$
VSAQ-9 : Find the value of log√2128
$$\text{Let } \log_{\sqrt{2}} 128 = x$$
$$(\sqrt{2})^x = 128$$
$$128 = 2^7$$
$$= (\sqrt{2})^{14}$$
$$\text{Bases are equal so powers should also be equal}$$
$$x = 14$$
$$\log_{\sqrt{2}} 128 = 14$$
VSAQ-10 : Write log 64/729 in its expansion form
$$\log \frac{64}{729} = \log 64 – \log 729$$
$$= \log 2^6 – \log 3^6$$
$$= 6 \log 2 – 6 \log 3$$
$$= 6(\log 2 – \log 3)$$
$$= 6 \log \left(\frac{2}{3}\right)$$