Class 9-10 Math Solution Exercise 4.2

Exercise 4.2: Logarithms

1. Find the values:

1) $\log_3 81$

Solution: $\log_3 81$

$= \log_3 (3^4)$

$= 4 \log_3 3$$[\because \log_a (M^r) = r \log_a M]$

$= 4 \times 1$$[\because \log_a a = 1]$

$= 4$

Answer:$4$

2) $\log_5 \sqrt[3]{5}$

Solution: $\log_5 \sqrt[3]{5}$

$= \log_5 (5^{\frac{1}{3}})$

$= \frac{1}{3} \log_5 5$

$= \frac{1}{3} \times 1 = \frac{1}{3}$

Answer:$\frac{1}{3}$

3) $\log_4 2$

Solution: $\log_4 2$

$= \log_4 (\sqrt{4})$

$= \log_4 (4^{\frac{1}{2}})$

$= \frac{1}{2} \log_4 4$

$= \frac{1}{2} \times 1 = \frac{1}{2}$

Answer:$\frac{1}{2}$

4) $\log_{2\sqrt{5}} 400$

Solution: $\log_{2\sqrt{5}} 400$

$= \log_{2\sqrt{5}} (2\sqrt{5})^4$$[\because (2\sqrt{5})^4 = 2^4 \times (\sqrt{5})^4 = 16 \times 25 = 400]$

$= 4 \log_{2\sqrt{5}} (2\sqrt{5})$

$= 4 \times 1 = 4$

Answer:$4$

5) $\log_5 (\sqrt[3]{5} \cdot \sqrt{5})$

Solution: $\log_5 (\sqrt[3]{5} \cdot \sqrt{5})$

$= \log_5 (5^{\frac{1}{3}} \cdot 5^{\frac{1}{2}})$

$= \log_5 (5^{\frac{1}{3} + \frac{1}{2}})$

$= \log_5 (5^{\frac{2+3}{6}})$

$= \log_5 (5^{\frac{5}{6}})$

$= \frac{5}{6} \log_5 5$

$= \frac{5}{6} \times 1 = \frac{5}{6}$

Answer:$\frac{5}{6}$

2. Find the value of $x$:

1) $\log_5 x = 3$

Solution: $\log_5 x = 3$

or, $x = 5^3$$[\because \log_a N = x \iff a^x = N]$

or, $x = 125$

Answer:$125$

2) $\log_x 25 = 2$

Solution: $\log_x 25 = 2$

or, $x^2 = 25$

or, $x^2 = 5^2$

or, $x = 5$$[x > 0]$

Answer:$5$

3) $\log_x \frac{1}{16} = -2$

Solution: $\log_x \frac{1}{16} = -2$

or, $x^{-2} = \frac{1}{16}$

or, $\frac{1}{x^2} = \frac{1}{16}$

or, $x^2 = 16$

or, $x = \sqrt{16}$

or, $x = 4$

Answer:$4$

3. Show that:

1) $5\log_{10}5 – \log_{10}25 = \log_{10}125$

Solution:

L.H.S. $= 5\log_{10}5 – \log_{10}25$

$= \log_{10}(5^5) – \log_{10}(5^2)$

$= \log_{10}\left(\frac{5^5}{5^2}\right)$

$= \log_{10}(5^{5-2})$

$= \log_{10}(5^3)$

$= \log_{10}125 = \text{R.H.S.}$

$\therefore 5\log_{10}5 – \log_{10}25 = \log_{10}125$ (Shown)

2) $\log_{10}\frac{50}{147} = \log_{10}2 + 2\log_{10}5 – \log_{10}3 – 2\log_{10}7$

Solution:

L.H.S. $= \log_{10}\frac{50}{147}$

$= \log_{10}50 – \log_{10}147$

$= \log_{10}(2 \times 5^2) – \log_{10}(3 \times 7^2)$

$= (\log_{10}2 + \log_{10}5^2) – (\log_{10}3 + \log_{10}7^2)$

$= \log_{10}2 + 2\log_{10}5 – \log_{10}3 – 2\log_{10}7 = \text{R.H.S.}$

$\therefore \log_{10}\frac{50}{147} = \log_{10}2 + 2\log_{10}5 – \log_{10}3 – 2\log_{10}7$ (Shown)

3) $3\log_{10}2 + 2\log_{10}3 + \log_{10}5 = \log_{10}360$

Solution:

L.H.S. $= 3\log_{10}2 + 2\log_{10}3 + \log_{10}5$

$= \log_{10}(2^3) + \log_{10}(3^2) + \log_{10}5$

$= \log_{10}8 + \log_{10}9 + \log_{10}5$

$= \log_{10}(8 \times 9 \times 5)$

$= \log_{10}360 = \text{R.H.S.}$

$\therefore 3\log_{10}2 + 2\log_{10}3 + \log_{10}5 = \log_{10}360$ (Shown)

4. Simplify:

1) $7\log_{10}\frac{10}{9} – 2\log_{10}\frac{25}{24} + 3\log_{10}\frac{81}{80}$

Solution: $7\log_{10}\frac{10}{9} – 2\log_{10}\frac{25}{24} + 3\log_{10}\frac{81}{80}$

$= \log_{10}\left(\frac{10}{9}\right)^7 – \log_{10}\left(\frac{25}{24}\right)^2 + \log_{10}\left(\frac{81}{80}\right)^3$

$= \log_{10}\left(\frac{2 \times 5}{3^2}\right)^7 – \log_{10}\left(\frac{5^2}{2^3 \times 3}\right)^2 + \log_{10}\left(\frac{3^4}{2^4 \times 5}\right)^3$

$= \log_{10}\left( \frac{2^7 \times 5^7}{3^{14}} \div \frac{5^4}{2^6 \times 3^2} \times \frac{3^{12}}{2^{12} \times 5^3} \right)$

$= \log_{10}\left( \frac{2^7 \times 5^7}{3^{14}} \times \frac{2^6 \times 3^2}{5^4} \times \frac{3^{12}}{2^{12} \times 5^3} \right)$

$= \log_{10}\left( \frac{2^{7+6} \times 3^{2+12} \times 5^7}{2^{12} \times 3^{14} \times 5^{4+3}} \right)$

$= \log_{10}\left( \frac{2^{13} \times 3^{14} \times 5^7}{2^{12} \times 3^{14} \times 5^7} \right)$

$= \log_{10}(2^{13-12})$

$= \log_{10}2$

Answer:$\log_{10}2$

2) $\log_7(\sqrt[5]{7} \cdot \sqrt{7}) – \log_3\sqrt[3]{3} + \log_4 2$

Solution: $\log_7(\sqrt[5]{7} \cdot \sqrt{7}) – \log_3\sqrt[3]{3} + \log_4 2$

$= \log_7(7^{\frac{1}{5}} \cdot 7^{\frac{1}{2}}) – \log_3(3^{\frac{1}{3}}) + \log_4(\sqrt{4})$

$= \log_7(7^{\frac{1}{5} + \frac{1}{2}}) – \frac{1}{3}\log_3 3 + \log_4(4^{\frac{1}{2}})$

$= \log_7(7^{\frac{7}{10}}) – \frac{1}{3}(1) + \frac{1}{2}\log_4 4$

$= \frac{7}{10}\log_7 7 – \frac{1}{3} + \frac{1}{2}(1)$

$= \frac{7}{10} – \frac{1}{3} + \frac{1}{2}$

$= \frac{21 – 10 + 15}{30}$

$= \frac{26}{30} = \frac{13}{15}$

Answer:$\frac{13}{15}$

3) $\log_e\frac{a^3 b^3}{c^3} + \log_e\frac{b^3 c^3}{d^3} + \log_e\frac{c^3 d^3}{a^3} – 3\log_e b^2 c$

Solution: $\log_e\frac{a^3 b^3}{c^3} + \log_e\frac{b^3 c^3}{d^3} + \log_e\frac{c^3 d^3}{a^3} – 3\log_e b^2 c$

$= \log_e\left( \frac{a^3 b^3}{c^3} \times \frac{b^3 c^3}{d^3} \times \frac{c^3 d^3}{a^3} \right) – \log_e (b^2 c)^3$

$= \log_e(b^6 c^3) – \log_e(b^6 c^3)$

$= 0$

Answer:$0$

5. $x = 2, y = 3, z = 5, w = 7$

1) What is the log of $\sqrt{y^3}$ to the base $3$.

Solution:

When $y = 3$, $\sqrt{y^3} = \sqrt{3^3} = 3^{\frac{3}{2}}$

Therefore, we get

$log_3\sqrt{y^3}$

$= \log_3 (3)^\frac{3}{2} $

$= \frac{3}{2} \log_3 3 = \frac{3}{2} \times 1 = \frac{3}{2}$

Answer:$\frac{3}{2}$

2) Find the value of $w\log\frac{xz}{y^2} – x\log\frac{z^2}{x^2 y} + y\log\frac{y^4}{x^4 z}$.

Solution:

Given, $x=2, y=3, z=5, w=7$

Given expression $= w\log\frac{xz}{y^2} – x\log\frac{z^2}{x^2 y} + y\log\frac{y^4}{x^4 z}$

$= 7\log\frac{2 \times 5}{3^2} – 2\log\frac{5^2}{2^2 \times 3} + 3\log\frac{3^4}{2^4 \times 5}$

$= \log_{10}\left(\frac{10}{9}\right)^7 – \log_{10}\left(\frac{25}{24}\right)^2 + \log_{10}\left(\frac{81}{80}\right)^3$

$= \log_{10}\left(\frac{2 \times 5}{3^2}\right)^7 – \log_{10}\left(\frac{5^2}{2^3 \times 3}\right)^2 + \log_{10}\left(\frac{3^4}{2^4 \times 5}\right)^3$

$= \log_{10}\left( \frac{2^7 \times 5^7}{3^{14}} \div \frac{5^4}{2^6 \times 3^2} \times \frac{3^{12}}{2^{12} \times 5^3} \right)$

$= \log_{10}\left( \frac{2^7 \times 5^7}{3^{14}} \times \frac{2^6 \times 3^2}{5^4} \times \frac{3^{12}}{2^{12} \times 5^3} \right)$

$= \log_{10}\left( \frac{2^{7+6} \times 3^{2+12} \times 5^7}{2^{12} \times 3^{14} \times 5^{4+3}} \right)$

$= \log_{10}\left( \frac{2^{13} \times 3^{14} \times 5^7}{2^{12} \times 3^{14} \times 5^7} \right)$

$= \log_{10}(2^{13-12})$

$= \log_{10}2$

Answer:$\log_{10}2$

3) Show that, $\frac{\log\sqrt{y^3} + y\log x – \frac{y}{x}\log(xz)}{\log(xy) – \log z} = \log_y\sqrt{y^3}$

Solution:

Substituting $x=2, y=3, z=5$:

R.H.S. $= \log_y\sqrt{y^3} = \log_3 (3^{\frac{3}{2}}) = \frac{3}{2} \log_3 3 = \frac{3}{2}$

L.H.S. $= \frac{\log\sqrt{3^3} + 3\log 2 – \frac{3}{2}\log(2 \times 5)}{\log(2 \times 3) – \log 5}$

$= \frac{\log (3^{\frac{3}{2}}) + \log (2^3) – \frac{3}{2}\log 10}{\log 6 – \log 5}$

$= \frac{\frac{3}{2}\log 3 + 3\log 2 – \frac{3}{2}(\log 2 + \log 5)}{\log\frac{6}{5}}$

$= \frac{\frac{3}{2}\log 3 + 3\log 2 – \frac{3}{2}\log 2 – \frac{3}{2}\log 5}{\log\frac{6}{5}}$

$= \frac{\frac{3}{2}\log 3 + \frac{3}{2}\log 2 – \frac{3}{2}\log 5}{\log\frac{6}{5}}$

$= \frac{\frac{3}{2}(\log 3 + \log 2 – \log 5)}{\log\left(\frac{2 \times 3}{5}\right)}$

$= \frac{\frac{3}{2}\log\left(\frac{6}{5}\right)}{\log\left(\frac{6}{5}\right)} = \frac{3}{2} = \text{R.H.S.}$

$\therefore \frac{\log\sqrt{y^3} + y\log x – \frac{y}{x}\log(xz)}{\log(xy) – \log z} = \log_y\sqrt{y^3}$ (Shown)

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