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Logarithms Made Super Easy (With Formulae & a Confidence-Boosting Test!)

Logarithms are essential for students due to their foundational role in higher mathematics, their ability to simplify complex calculations, and their widespread applications in science, technology, and problem-solving. Get easy-to-understand theory, Formulae, and Interactive Exam-Style Questions with Instant Scoring and Detailed Explanations.
Logarithms Made Super Easy (With Formulae & a Confidence-Boosting Test!)

What are Logarithms?

Logarithms trying to find out the power you have to raise to get figure out another number.

Logarithm Formulae and Practice Questions

Logarithm Formulae and Practice Questions

The main Logarithm Formula is:


If \( b^x \) = n, then \( \log_{b}(n) = x \)

Common Logarithm Formulas:

\( \log_{b}(1) = 0 \)

\( \log_{b}(b) = 1 \)

\( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \) (Product Rule)

\( \log_{b}(x/y) = \log_{b}(x) - \log_{b}(y) \) (Quotient Rule)

\( \log_{b}(x^n) = n*\log_{b}(x) \) (Quotient Rule)

\( \log_{b}(a) = \log_{c}(a) / \log_{c}(b) \) (Change of Base Formula)



Logarithm Quiz


Please Note: It is difficult to write log answers in the check box. Hence use the following syntax :

\( 4 \log_{b}(x^n) \) is written as 4*logb(x^n) ; Use sqrt for square root and +- for ±


1. Solve for \( x \) in the following equations:

a) \( \log_{4}(x) = 2 \)

\( \log_{4}(x) = 2 \implies x = 4^2 = 16 \)

b) \( \log_{5}(x) = 3 \)

\( \log_{5}(x) = 3 \implies x = 5^3 = 125 \)

2. Evaluate the following logarithms:

a) \( \log_{2}(16) \)

\( \log_{2}(16) = 4 \)

b) \( \log_{3}(81) \)

\( \log_{3}(81) = 4 \)

3. Simplify the following logarithms:

a) \( \log_{5}(25) \)

\( \log_{5}(25) = 2 \)

b) \( \log_{2}\left(\frac{1}{8}\right) \)

\( \log_{2}\left(\frac{1}{8}\right) = -3 \)

4. Use the change of base formula to evaluate:

a) \( \log_{4}(16) \)

\( \log_{4}(16) = 2 \)

b) \( \log_{6}(36) \)

\( \log_{6}(36) = 2 \)

5. Solve for \( x \) in the following logarithmic equations:

a) \( \log_{3}(x) = 1 + \log_{3}(27) \)

\( \log_{3}(x) = 1 + \log_{3}(27) \implies log_{3}(27)=3 \implies \log_{3}(x) = 1 + 3 \implies 3^4=x \implies x=81 \)

b) \( \log_{2}(2x) = 3 \)

\( \log_{2}(2x) = 3 \implies 2x = 2^3 = 8 \implies x = 4 \)

6. Expand the following logarithms:

a) \( \log_{7}(x^2 y) \)

\( \log_{7}(x^2 y) = 2 \log_{7}(x) + \log_{7}(y) \)

b) \( \log_{2}\left(\frac{x}{y^3}\right) \)

\( \log_{2}\left(\frac{x}{y^3}\right) = \log_{2}(x) - 3 \log_{2}(y) \)

7. Condense the following logarithms:

a) \( 3 \log_{2}(x) - \log_{2}(y) \)

\( 3 \log_{2}(x) - \log_{2}(y) = \log_{2}(x^3) - \log_{2}(y) \) = \( \log_{2}\left(\frac{x^3}{y}\right) \)

b) \( \frac{1}{2} \log_{3}(x) + \log_{3}(y) \)

\( \frac{1}{2} \log_{3}(x) + \log_{3}(y) = \log_{3}(\sqrt{x}) + \log_{3}(y) = \log_{3}\left(\frac{\sqrt{x}}{y}\right) \)

8. Solve for \( x \) in the following logarithmic equations:

a) \( \log_{4}(x-1) + \log_{4}(x+1) = 2 \)

\( \log_{4}(x-1) + \log_{4}(x+1) = 2 \implies (x-1)(x+1) = 16 \implies x^2 - 1 = 16 \implies x^2 = 17 \implies x = \pm\sqrt{17} \)

b) \( \log_{3}(x+2) - \log_{3}(x-2) = 2 \)

\( \log_{3}(x+2) - \log_{3}(x-2) = 2 \implies \log_{3}\left(\frac{x+2}{x-2}\right) = 2 \implies \frac{x+2}{x-2} = 3^2 \implies x + 2 = 9(x - 2) \implies x = 2.5 \)

9. Evaluate the following logarithms using properties:

a) \( \log_{5}(125) \)

\( \log_{5}(125) = 3 \)

b) \( \log_{3}(27) \)

\( \log_{3}(27) = 3 \)

10. Solve for \( x \) in the following equations:

a) \( \log_{2}(x) + \log_{2}(x-2) = 3 \)

\( \log_{2}(x) + \log_{2}(x-2) = 3 \implies \log_{2}(x(x-2)) = 3 \implies x(x-2) = 2^3 = 8 \implies x^2 - 2x - 8 = 0 \implies x = 4 \)
Note: The logarithm log⁡2(x) is undefined for negative values since logarithms are only defined for positive arguments. Therefore, x=−2 is not a valid solution in this context.

b) \( \log_{3}(x^2 - 4) = 2 \)

\( \log_{3}(x^2 - 4) = 2 \implies x^2 - 4 = 3^2 = 9 \implies x^2 = 13 \implies x = \pm\sqrt{13} \)

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