Intuition The big pictureA logarithmic equation hides an unknown inside a log, like log 2 ( x ) = 5 \log_2(x) = 5 log 2 ( x ) = 5 undo the logarithm . The single most powerful idea is:
log b ( a ) = c ⟺ b c = a \log_b(a)=c \iff b^c = a log b ( a ) = c ⟺ b c = a asks a question : "log 2 8 \log_2 8 log 2 8 "2 to what power gives 8?" Solving log equations = rewriting that question as its exponential answer, or collapsing many logs into one so we can undo them.
Definition Logarithmic equationAn equation where the unknown appears in the argument (or occasionally the base) of a logarithm, e.g. log 3 ( 2 x − 1 ) = 4 \log_3(2x-1)=4 log 3 ( 2 x − 1 ) = 4 x x x both sides are defined and equal .
Crucial restriction (domain): log b ( N ) \log_b(N) log b ( N )
the argument N > 0 N>0 N > 0 0 0 0
the base b > 0 b>0 b > 0 b ≠ 1 b\ne 1 b = 1
So every solution MUST be checked against N > 0 N>0 N > 0
WHY it works: raising the base to each side "cancels" the log because b log b N = N b^{\log_b N}=N b l o g b N = N
log b ( N ) = c ⟹ b log b N = b c ⟹ N = b c \log_b(N)=c \;\Longrightarrow\; b^{\log_b N}=b^c \;\Longrightarrow\; N=b^c log b ( N ) = c ⟹ b l o g b N = b c ⟹ N = b c
WHY: the log function is one-to-one (strictly increasing/decreasing), so
log b M = log b N ⟹ M = N ( M , N > 0 ) . \log_b M=\log_b N \;\Longrightarrow\; M=N \quad(M,N>0). log b M = log b N ⟹ M = N ( M , N > 0 ) .
Worked example Example 1 — single logSolve log 2 ( 3 x − 1 ) = 4 \log_2(3x-1)=4 log 2 ( 3 x − 1 ) = 4
Step 1. Argument is already isolated on the left as one log. Why? So we can exponentiate directly.
Step 2. Rewrite in exponential form: 3 x − 1 = 2 4 = 16 3x-1=2^4=16 3 x − 1 = 2 4 = 16 Why? log b N = c ⟺ N = b c \log_b N=c\iff N=b^c log b N = c ⟺ N = b c Step 3. Solve: 3 x = 17 ⇒ x = 17 3 3x=17\Rightarrow x=\tfrac{17}{3} 3 x = 17 ⇒ x = 3 17 Step 4 (check domain). Argument = 3 ( 17 3 ) − 1 = 16 > 0 =3(\tfrac{17}{3})-1=16>0 = 3 ( 3 17 ) − 1 = 16 > 0 Why? A log's argument must be positive.
x = 17 3 \boxed{x=\tfrac{17}{3}} x = 3 17
Worked example Example 2 — combine two logsSolve log 5 ( x ) + log 5 ( x − 4 ) = 1 \log_5(x)+\log_5(x-4)=1 log 5 ( x ) + log 5 ( x − 4 ) = 1
Step 1. Combine using the product law: log 5 ( x ( x − 4 ) ) = 1 \log_5\big(x(x-4)\big)=1 log 5 ( x ( x − 4 ) ) = 1 Why? Two separate logs can't be undone; one log can.
Step 2. Exponentiate: x ( x − 4 ) = 5 1 = 5 x(x-4)=5^1=5 x ( x − 4 ) = 5 1 = 5 Why? log 5 N = 1 ⟺ N = 5 \log_5 N=1\iff N=5 log 5 N = 1 ⟺ N = 5 Step 3. Expand: x 2 − 4 x − 5 = 0 ⇒ ( x − 5 ) ( x + 1 ) = 0 ⇒ x = 5 x^2-4x-5=0\Rightarrow(x-5)(x+1)=0\Rightarrow x=5 x 2 − 4 x − 5 = 0 ⇒ ( x − 5 ) ( x + 1 ) = 0 ⇒ x = 5 x = − 1 x=-1 x = − 1 Step 4 (check!). Need x > 0 x>0 x > 0 and x − 4 > 0 x-4>0 x − 4 > 0 x > 4 x>4 x > 4
x = 5 x=5 x = 5 5 > 0 5>0 5 > 0 1 > 0 1>0 1 > 0 x = − 1 x=-1 x = − 1 log 5 ( − 1 ) \log_5(-1) log 5 ( − 1 ) reject .
x = 5 \boxed{x=5} x = 5
Worked example Example 3 — logs on both sidesSolve log 3 ( 2 x + 1 ) = log 3 ( x + 7 ) \log_3(2x+1)=\log_3(x+7) log 3 ( 2 x + 1 ) = log 3 ( x + 7 )
Step 1. Same base both sides → equate arguments: 2 x + 1 = x + 7 2x+1=x+7 2 x + 1 = x + 7 Why? log \log log Step 2. x = 6 x=6 x = 6 Step 3 (check). 2 ( 6 ) + 1 = 13 > 0 2(6)+1=13>0 2 ( 6 ) + 1 = 13 > 0 6 + 7 = 13 > 0 6+7=13>0 6 + 7 = 13 > 0 x = 6 \boxed{x=6} x = 6
Worked example Example 4 — hidden quadratic in a logSolve ( log 2 x ) 2 − 3 log 2 x + 2 = 0 (\log_2 x)^2-3\log_2 x+2=0 ( log 2 x ) 2 − 3 log 2 x + 2 = 0
Step 1. Let u = log 2 x u=\log_2 x u = log 2 x Why? The equation is quadratic in the log , not in x x x Step 2. u 2 − 3 u + 2 = 0 ⇒ ( u − 1 ) ( u − 2 ) = 0 ⇒ u = 1 u^2-3u+2=0\Rightarrow(u-1)(u-2)=0\Rightarrow u=1 u 2 − 3 u + 2 = 0 ⇒ ( u − 1 ) ( u − 2 ) = 0 ⇒ u = 1 u = 2 u=2 u = 2 Step 3. Back-substitute: log 2 x = 1 ⇒ x = 2 \log_2 x=1\Rightarrow x=2 log 2 x = 1 ⇒ x = 2 log 2 x = 2 ⇒ x = 4 \log_2 x=2\Rightarrow x=4 log 2 x = 2 ⇒ x = 4 Step 4 (check). Both 2 > 0 , 4 > 0 2>0,4>0 2 > 0 , 4 > 0 x = 2 or x = 4 \boxed{x=2 \text{ or } x=4} x = 2 or x = 4
Common mistake Forgetting to check the domainWrong feels right because: algebra gave you a valid number, so surely it's a solution. But the original equation may be undefined there.
Fix: After solving, substitute back and confirm every log argument is > 0 >0 > 0
log ( A ) + log ( B ) = log ( A + B ) \log(A)+\log(B) = \log(A+B) log ( A ) + log ( B ) = log ( A + B ) Feels right because: it looks symmetric and "distributive". But logs turn products into sums, not sums into sums.
Fix: log A + log B = log ( A B ) \log A+\log B=\log(AB) log A + log B = log ( A B ) no rule for log ( A + B ) \log(A+B) log ( A + B )
Common mistake Cancelling a log that equals a number by "equating arguments"On log 5 x + log 5 ( x − 4 ) = 1 \log_5 x+\log_5(x-4)=1 log 5 x + log 5 ( x − 4 ) = 1 x + ( x − 4 ) = 1 x+(x-4)=1 x + ( x − 4 ) = 1 Feels right because: you can equate when both sides are single logs.
Fix: The right side is a number , not a log. Combine first, then exponentiate (= 5 1 =5^1 = 5 1
Recall Feynman: explain to a 12-year-oldA logarithm is a "power detective." log 2 8 \log_2 8 log 2 8 "how many 2's do I multiply to get 8?" — answer 3. So a log equation is a riddle with a missing number inside the detective's question. To solve it you flip the riddle around: instead of "what power gives this number?", you say "the number IS the base raised to that power," and now you can just do normal algebra. Golden rule: at the end, plug your answer back — because you're never allowed to ask "what power gives a negative number?" with a positive base, so any answer that makes the inside negative is a trick and gets thrown out.
Mnemonic Remember the workflowC.E.C. → C ombine, E xponentiate, C heck.
("Cats Eat Cheese " — always finish by Check ing the domain, or the cat goes hungry.)
What does log b N = c \log_b N = c log b N = c Why must you always check solutions of a log equation? Because log arguments must be
> 0 >0 > 0 ; algebra can produce extraneous roots that make an argument
≤ 0 \le 0 ≤ 0 .
log a M + log a N = ? \log_a M + \log_a N = ? log a M + log a N = ? log a ( M N ) \log_a(MN) log a ( M N ) (product law)
log a M − log a N = ? \log_a M - \log_a N = ? log a M − log a N = ? log a ( M / N ) \log_a(M/N) log a ( M / N ) (quotient law)
k log a M = ? k\log_a M = ? k log a M = ? log a ( M k ) \log_a(M^k) log a ( M k ) (power law)
When can you equate arguments directly? When both sides are single logs of the same base:
log b M = log b N ⇒ M = N \log_b M=\log_b N\Rightarrow M=N log b M = log b N ⇒ M = N .
Solve log 2 ( 3 x − 1 ) = 4 \log_2(3x-1)=4 log 2 ( 3 x − 1 ) = 4 3 x − 1 = 16 ⇒ x = 17 / 3 3x-1=16\Rightarrow x=17/3 3 x − 1 = 16 ⇒ x = 17/3 .
Why does b log b N = N b^{\log_b N}=N b l o g b N = N Because
log b \log_b log b and
b ( ⋅ ) b^{(\cdot)} b ( ⋅ ) are inverse functions — one undoes the other.
Is log ( A ) + log ( B ) = log ( A + B ) \log(A)+\log(B)=\log(A+B) log ( A ) + log ( B ) = log ( A + B ) No — it equals
log ( A B ) \log(AB) log ( A B ) ; there's no rule for
log ( A + B ) \log(A+B) log ( A + B ) .
For ( log x ) 2 − 3 log x + 2 = 0 (\log x)^2-3\log x+2=0 ( log x ) 2 − 3 log x + 2 = 0 Let
u = log x u=\log x u = log x , giving a quadratic
u 2 − 3 u + 2 = 0 u^2-3u+2=0 u 2 − 3 u + 2 = 0 .
Log equation: unknown in argument
Domain: argument N>0, base b>0 not 1
Strategy 1: Isolate and exponentiate
Strategy 2: Combine logs then exponentiate
Strategy 3: Same base, equate arguments
Index laws b^x b^y = b^x+y
Log laws: product, quotient, power
Intuition Hinglish mein samjhoDekho, logarithmic equation ka matlab hai ki unknown x x x andar chhupa hua hai, jaise log 2 ( 3 x − 1 ) = 4 \log_2(3x-1)=4 log 2 ( 3 x − 1 ) = 4 log b N = c \log_b N = c log b N = c N = b c N = b^c N = b c 2 2 2 8 8 8
Teen main tricks hain. Ek: agar ek hi log hai, to use isolate karke base ki power utha do (exponentiate). Do: agar do-teen log alag-alag hain, to pehle log laws (log M + log N = log M N \log M+\log N=\log MN log M + log N = log M N ek log me combine karo, phir exponentiate karo — kyunki do alag log ek saath undo nahi hote. Teen: agar dono side par same base ke log hain, to seedha arguments equal kar do, kyunki log function one-to-one hota hai.
Sabse bada exam trap: domain check bhoolna . Log ka argument hamesha positive (> 0 >0 > 0
Yaad rakhne ka mantra: C.E.C. — Combine, Exponentiate, Check. Bas yeh order follow karo aur har log equation crack ho jayegi. Check wala step chhota lagta hai par wahi marks bachata hai!