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Elite lesson· 31 min English Unlock every lessonFormulas At a glance 13 formulas
1 Definition log a x = y ⟺ a y = x \log_a x = y \iff a^y = x log a x = y ⟺ a y = x a > 0 , a ≠ 1 , x > 0 a > 0,\ a \ne 1,\ x > 0 a > 0 , a = 1 , x > 0 2 Values to know log a 1 = 0 , log a a = 1 \log_a 1 = 0,\quad \log_a a = 1 log a 1 = 0 , log a a = 1 a log a x = x a^{\log_a x} = x a l o g a x = x 3 Product and power log m n = log m + log n \log mn = \log m + \log n log mn = log m + log n log m k = k log m \log m^{k} = k\log m log m k = k log m 4 Quotient log m n = log m − log n \log \dfrac{m}{n} = \log m - \log n log n m = log m − log n log ( m + n ) ≠ log m + log n \log (m + n) \ne \log m + \log n log ( m + n ) = log m + log n 5 Change of base log a b = log b log a \log_a b = \dfrac{\log b}{\log a} log a b = log a log b log a b × log b a = 1 \log_a b \times \log_b a = 1 log a b × log b a = 1 6 Power in the base log a k x = 1 k log a x \log_{a^k} x = \dfrac{1}{k}\,\log_a x log a k x = k 1 log a x log a k x m = m k log a x \log_{a^k} x^{m} = \dfrac{m}{k}\,\log_a x log a k x m = k m log a x 7 Chain of logs log a b × log b c × log c d = log a d \log_a b \times \log_b c \times \log_c d = \log_a d log a b × log b c × log c d = log a d 8 The flip a log b c = c log b a a^{\log_b c} = c^{\log_b a} a l o g b c = c l o g b a 9 Nested logs Peel from the outside with log a y = k ⇒ y = a k \log_a y = k \Rightarrow y = a^k log a y = k ⇒ y = a k : log 2 ( log 3 y ) = 1 \log_2(\log_3 y) = 1 log 2 ( log 3 y ) = 1 gives log 3 y = 2 \log_3 y = 2 log 3 y = 2 , so y = 9 y = 9 y = 9 . 10 Check the domain Every log needs a positive input and a positive base other than 1. Solve first, then throw out any root that makes an input 0 or negative. 11 Log inequality a > 1 : log a x > log a y ⟺ x > y > 0 a > 1:\ \log_a x > \log_a y \iff x > y > 0 a > 1 : log a x > log a y ⟺ x > y > 0 0 < a < 1 : log a x > log a y ⟺ 0 < x < y 0 < a < 1:\ \log_a x > \log_a y \iff 0 < x < y 0 < a < 1 : log a x > log a y ⟺ 0 < x < y 12 Quadratic in a log With log a x \log_a x log a x and log x a \log_x a log x a together, put t = log a x t = \log_a x t = log a x , so log x a = 1 t \log_x a = \frac{1}{t} log x a = t 1 . If the roots are t 1 , t 2 t_1, t_2 t 1 , t 2 , the solutions x x x multiply to a t 1 + t 2 a^{t_1 + t_2} a t 1 + t 2 . 13 Number of digits digits in N = ⌊ log 10 N ⌋ + 1 \text{digits in } N = \lfloor \log_{10} N \rfloor + 1 digits in N = ⌊ log 10 N ⌋ + 1 whole N ≥ 1 ; e.g. log 10 2 50 ≈ 15.05 , so 16 digits \text{whole } N \ge 1;\ \text{e.g. } \log_{10} 2^{50} \approx 15.05,\ \text{so 16 digits} whole N ≥ 1 ; e.g. log 10 2 50 ≈ 15.05 , so 16 digits Test yourself Q1 Easy
If log x 64 = 3 \log_x 64 = 3 log x 64 = 3 , what is x x x ?
A 4 B 8 C 64 3 \frac{64}{3} 3 64 D 192
Correct answer· 4
Why
Turn the log into a power: log x 64 = 3 \log_x 64 = 3 log x 64 = 3 means x 3 = 64 x^3 = 64 x 3 = 64 .
4 3 = 64 4^3 = 64 4 3 = 64 ; taking the square root gives the trap 8, and dividing 64 by 3 gives 64 3 \frac{64}{3} 3 64 .
So x = 4 x = 4 x = 4 .
Q2 Easy
Solve for x x x : log 2 ( x − 3 ) + log 2 ( x + 1 ) = 5 \log_2(x - 3) + \log_2(x + 1) = 5 log 2 ( x − 3 ) + log 2 ( x + 1 ) = 5 .
A 5 B 6 C 7 D 8
Correct answer· 7
Why
Join the two logs into one, then rewrite it as a power of 2.
( x − 3 ) ( x + 1 ) = 2 5 = 32 (x - 3)(x + 1) = 2^5 = 32 ( x − 3 ) ( x + 1 ) = 2 5 = 32 means x 2 − 2 x − 35 = 0 x^2 - 2x - 35 = 0 x 2 − 2 x − 35 = 0 , whose roots are 7 and − 5 -5 − 5 .
Only 7 keeps both inputs positive, so x = 7 x = 7 x = 7 .
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