### Shortcut Methods

**JEE Main**

**Find the value of log10(1024) - log10(0.0625).**

**Shortcut method:**

- Write 1024 as 10×10×10.
- Write 0.0625 as 1/16.
- Now, log10(1024) - log10(0.0625) = log10(10×10×10) - log10(1/16) = 3 log10(10) - log10(1) + log10(16) = 3 - 0 + 4 = 7

**Trick:**

- Notice that 1024 = 2^10 and 0.0625 = (1/4)^4.
- Therefore, log10(1024) - log10(0.0625) = log10(2^10) - log10((1/4)^4) = 10 log10(2) - 4 log10(1/4) = 10 - 4(-2) = 18.

**Solve the equation 2log2x = log216.**

**Shortcut method:**

- Divide both sides of the equation by log2: 2log2x/log2 = log216/log2 ⇒ 2 = 4

This is not true, so the equation has no solution.

**Trick:**

- Notice that 2log2x = log2(2^2x) and log216 = log2(6^2).
- Therefore, the equation 2log2x = log216 is equivalent to 2^2x = 6^2.
- Solving this equation gives x = 3.

**Find the domain and range of the function log2(3-x).**

**Shortcut method:**

- The argument of the logarithm, 3-x, must be greater than 0 for the logarithm to be defined.
- Therefore, the domain of the function is {x | x < 3}.
- The range of the function is all real numbers since the logarithm function can take on any real value.

**Trick:**

- Notice that the function log2(3-x) is a decreasing function.
- Therefore, the range of the function is the interval (-∞, log23).

**JEE Advanced**

**Find the value of log10(2018!) - log10(2017!).**

**Shortcut method:**

- Use the fact that log10(n!) = log10(1×2×3…n) = log10(1) + log10(2) + log10(3) + … + log10(n).
- Therefore, log10(2018!) - log10(2017!) = [log10(1) + log10(2) + log10(3) + … + log10(2018)] - [log10(1) + log10(2) + log10(3) + … + log10(2017)] = log10(2018).

**Trick:**

- Notice that the sum of the first 2017 positive integers is given by the formula: 1+2+3+4+…+2017= 2017×(2018)/2 = 2035896.
- Therefore, log10(2018!) - log10(2017!) = log10(2018) - log10(2035896) = log10(2018/2035896) = log10(1/1016) = -3.

**Solve the equation log2(x+2) + log2(x-2) = 3.**

**Shortcut method:**

- Use the product rule of logarithms to combine the two logarithms on the left-hand side of the equation: log2(x+2) + log2(x-2) = log2((x+2)(x-2)) = 3.
- Now, solve the quadratic equation (x+2)(x-2) = 2^3: x^2 - 4 = 8 x^2 = 12 x = ±√12 = ±2√3

**Trick:**

- Notice that the expression (x+2)(x-2) is positive when x > 2 and negative when x < 2.
- Therefore, the equation log2(x+2) + log2(x-2) = 3 has two solutions, x = 2 + 2√3 and x = 2 - 2√3.

**Find the area of the region bounded by the curves y = log2(x), y = 0, and x = 4.**

**Shortcut method:**

- The area of the region bounded by the curves y = log2(x), y = 0, and x = 4 is given by the integral of log2(x) with respect to x, from x = 1 to x = 4: ∫1^4 log2(x) dx
- Use u-substitution with u = log2(x). Then du = (1/x ln2) dx and x = 2^u.
- Substituting into the integral, we get: ∫1^4 log2(x) dx = ∫0^2 u (2^u ln2) du = ln2 ∫0^2 u 2^u du = ln2 [u 2^u - 2^u]0^2 = ln2 (4 - 4) = 0.

**Trick:**

- Notice that the graph of the function y = log2(x) is a concave up function.
- Therefore, the area of the region bounded by the curves y = log2(x), y = 0, and x = 4 is equal to the area of the triangle formed by the points (1, 0), (4, 2), and (4, 0).
- This area is given by the formula: Area = ½ x base x height = ½ x 3 x 2 = 3 square units.

**CBSE Board Exams**

**Find the value of log10(100) + log10(0.01).**

**Shortcut method:**

- Use the laws of logarithms to simplify the expression: log10(100) + log10(0.01) = log10(100 × 0.01) = log10(1) = 0.

**Trick:**

- Notice that 100 = 10^2 and 0.01 = 10^-2.
- Therefore, log10(100) + log10(0.01) = log10(10^2) + log10(10^-2) = 2 - (-2) = 4.

**Solve the equation logx25 = 2.**

**Shortcut method:**

- Use the definition of logarithms to rewrite the equation as an exponential equation: logx25 = 2 ⇔ 25 = x^2.

-Now, solve the quadratic equation x^2 = 25: x = ±√25 = ±5. -Therefore, x = 5 is the only solution to the equation logx25 = 2.

**Trick:**

- Notice that 25 is the square of 5. -Therefore, the equation logx25 = 2 is equivalent to the equation x = 5.

**Find the domain and range of the function logy(2x-1).**

**Shortcut method:**

- The argument of the logarithm, 2x-1, must be greater than 0 for the logarithm to be defined. -Therefore, the domain of the function is {x | x > 1/2}.
- The range of the function is all real numbers since the logarithm function can take on any real value.

**Trick:**

- Notice that the function logy(2x-1) is an increasing function. -Therefore, the range of the function is the interval (-∞, ∞).