JEE Mains

1. Express the product of (5^{\sqrt{3}}) and (5^{-\frac{1}{2}}) in the form of (5^x), where x is a rational number.

Shortcut Method: Use the laws of exponents: $$5^{\sqrt{3}}\times5^{-\frac{1}{2}}=5^{\sqrt{3}-\frac{1}{2}}=5^{\frac{2\sqrt{3}-1}{2}}$$ Therefore, (x = \frac{2\sqrt{3}-1}{2}).

2. Find the value of (\log_{2} 32 - 2\log_{2} 8 + \log_{2} 4).

Shortcut Method: Use the laws of logarithms: (\log_{2} 32 - 2\log_{2} 8 + \log_{2} 4) (= \log_{2} 2^5 - 2\log_{2} 2^3 + \log_{2} 2^2) (= \log_{2} 2^2) (= 2).

3. Simplify (\log_{10} \frac{1}{100} + \log_{10} 100 - \log_{10} 10).

Shortcut Method: Use the laws of logarithms: (\log_{10} \frac{1}{100} + \log_{10} 100 - \log_{10} 10) (= \log_{10} 100\cdot\frac{1}{100} + \log_{10} 100 - \log_{10} 10) (= \log_{10} 1 + \log_{10} 100 - \log_{10} 10) (= 0).

4. If (\log_{a} b = 2) and (\log_{a} c = 3), find the value of (\log_{a} \frac{b^3c^2}{a^5}).

Shortcut Method: Use the laws of logarithms: (\log_{a} \frac{b^3c^2}{a^5}) (= \log_{a} b^3 + \log_{a} c^2 - \log_{a} a^5) (= 3\log_{a} b + 2\log_{a} c - 5\log_{a} a) (= 3(2) + 2(3) - 5(1)) (= 6 + 6 - 5). (= 7).

5. Show that (\log_{a} b = \frac{1}{\log_{b} a}) for any positive numbers a and b.

Proof: Let (x = \log_{a} b). Then, (a^x = b). Taking logarithms on both sides of the equation with base b, we get: (\log_{b} a^x = \log_{b} b) (x\log_{b} a = 1) (\log_{a} b = \frac{1}{\log_{b} a}).

CBSE Board Exams

1. Find the value of (2\log_{3} 9 - 3\log_{3} 4).

Shortcut Method: Use the laws of logarithms: (2\log_{3} 9 - 3\log_{3} 4) (= \log_{3} 9^2 - \log_{3} 4^3) (= \log_{3} \frac{9^2}{4^3}) (= \log_{3} \frac{81}{64}) (= \log_{3} \left(\frac{3}{2}\right)^4) (= 4\log_{3} \left(\frac{3}{2}\right)) (= 4\left(\log_{3} 3 - \log_{3} 2\right)) (= 4(1 - \log_{3} 2)) (= 4 - 4\log_{3} 2).

2. Express the quotient of (\frac{\sqrt[3]{8}}{2\sqrt[4]{16}}) in the form of a single logarithm.

Shortcut Method: Use the laws of logarithms: (\frac{\sqrt[3]{8}}{2\sqrt[4]{16}}) (= \frac{2^{3/3}}{2\cdot 2^{4/4}}) (= \frac{2^{1}}{2^{1}}) (= 1) (\log_{a} b^c = c\log_{a} b) (\log_{2} 1 = 0).

3. Simplify (\log_{5} 25 - \log_{5} 5).

Shortcut Method: Use the laws of logarithms: (\log_{5} 25 - \log_{5} 5) (= \log_{5} \frac{25}{5}) (= \log_{5} 5) (= 1)

4. Find the value of (\log_{10} 0.0001).

Shortcut Method: (\log_{10} 0.0001=\log_{10}(10^{-4})) (\log_{10}0.0001=-4).

5. Show that (\log_{a} (ab) = 1 + \log_{a} b) for any positive numbers a and b.

Proof: Let (x = \log_{a} (ab)). Then, (a^x = ab). Dividing both sides of the equation by (a), we get: (\frac{a^x}{a} = \frac{ab}{a}) (a^{x-1} = b) Taking logarithms on both sides of the equation with base a, we get: (\log_{a} a^{x-1} = \log_{a} b) (x-1 = \log_{a} b) (\log_{a} (ab) = 1 + \log_{a} b).

Typical Numerical Values

• (\log_{10} 2 = 0.3010)
• (\log_{10} 3 = 0.4771)
• (\log_{10} 5 = 0.6990)
• (\log_{2} 10 = 3.3219)
• (\log_{2} 20 = 4.3219)