A logarithm asks a question : "log b x \log_b x log b x "to what power must I raise b b b x x x
Because logs are exponents in disguise , every rule for exponents becomes a rule for logs.
Multiplying numbers ⇄ adding their exponents. Dividing ⇄ subtracting exponents. Raising to a power ⇄ multiplying the exponent. That single mirror is the whole topic.
Definition Definition of logFor base b > 0 , b ≠ 1 b>0,\ b\neq 1 b > 0 , b = 1 x > 0 x>0 x > 0 log b x = y ⟺ b y = x \log_b x = y \quad \Longleftrightarrow \quad b^{y} = x log b x = y ⟺ b y = x b log b x = x and log b ( b k ) = k b^{\log_b x} = x \qquad\text{and}\qquad \log_b(b^{k}) = k b l o g b x = x and log b ( b k ) = k
WHY the restrictions? We need b > 0 b>0 b > 0 b ≠ 1 b\neq 1 b = 1 1 y = 1 1^y=1 1 y = 1 x > 0 x>0 x > 0 always positive — you can never take a log of 0 0 0
The trick every time: name the logs as exponents , use exponent laws, then translate back.
log b ( M N ) = log b M + log b N \log_b(MN)=\log_b M+\log_b N log b ( M N ) = log b M + log b N Step 1. Let p = log b M p=\log_b M p = log b M q = log b N q=\log_b N q = log b N Why? Give the unknown logs names so we can work with them as exponents.
Step 2. By the definition, b p = M b^{p}=M b p = M b q = N b^{q}=N b q = N Why? This is exactly what "log \log log means — undo the log into exponential form.
Step 3. Multiply: M N = b p ⋅ b q = b p + q MN = b^{p}\cdot b^{q} = b^{p+q} M N = b p ⋅ b q = b p + q Why? Same-base multiplication adds exponents (b p b q = b p + q b^{p}b^{q}=b^{p+q} b p b q = b p + q
Step 4. Take log b \log_b log b log b ( M N ) = log b ( b p + q ) = p + q \log_b(MN)=\log_b(b^{p+q})=p+q log b ( M N ) = log b ( b p + q ) = p + q Why? Because log b ( b k ) = k \log_b(b^{k})=k log b ( b k ) = k
Step 5. Replace p , q p,q p , q log b ( M N ) = log b M + log b N . ■ \log_b(MN)=\log_b M+\log_b N.\ \blacksquare log b ( M N ) = log b M + log b N . ■
log b ( M / N ) = log b M − log b N \log_b(M/N)=\log_b M-\log_b N log b ( M / N ) = log b M − log b N Step 1–2. Same start: p = log b M , q = log b N p=\log_b M,\ q=\log_b N p = log b M , q = log b N b p = M , b q = N b^{p}=M,\ b^{q}=N b p = M , b q = N Step 3. Divide: M N = b p b q = b p − q \dfrac{M}{N}=\dfrac{b^{p}}{b^{q}}=b^{p-q} N M = b q b p = b p − q Why? Same-base division subtracts exponents .
Step 4. log b ( M N ) = log b ( b p − q ) = p − q \log_b\!\left(\tfrac{M}{N}\right)=\log_b(b^{p-q})=p-q log b ( N M ) = log b ( b p − q ) = p − q Step 5. = log b M − log b N . ■ =\log_b M-\log_b N.\ \blacksquare = log b M − log b N . ■
log b ( M k ) = k log b M \log_b(M^{k})=k\log_b M log b ( M k ) = k log b M Step 1. Let p = log b M p=\log_b M p = log b M b p = M b^{p}=M b p = M Step 2. Raise both sides to power k k k M k = ( b p ) k = b p k M^{k}=(b^{p})^{k}=b^{pk} M k = ( b p ) k = b p k Why? A power of a power multiplies exponents (( b p ) k = b p k (b^p)^k=b^{pk} ( b p ) k = b p k Step 3. log b ( M k ) = log b ( b p k ) = p k = k p \log_b(M^{k})=\log_b(b^{pk})=pk=kp log b ( M k ) = log b ( b p k ) = p k = k p Step 4. = k log b M . ■ =k\log_b M.\ \blacksquare = k log b M . ■
Bonus (reciprocal): put k = − 1 k=-1 k = − 1 log b ( 1 / M ) = − log b M \log_b(1/M)=-\log_b M log b ( 1/ M ) = − log b M
Intuition What earns 90% of the marks
Definition ⇄ exponential form (turn every log into b something b^{\text{something}} b something Multiply→add, divide→subtract, power→multiply-out-front. Combine and split expressions fluently — that is what exam questions test.
Worked example Combine into one logWrite 2 log 3 x + log 3 5 − log 3 y 2\log_3 x + \log_3 5 - \log_3 y 2 log 3 x + log 3 5 − log 3 y Step 1. 2 log 3 x = log 3 x 2 2\log_3 x = \log_3 x^{2} 2 log 3 x = log 3 x 2 Why? Power rule in reverse — bring the coefficient up as a power.
Step 2. log 3 x 2 + log 3 5 = log 3 ( 5 x 2 ) \log_3 x^{2}+\log_3 5 = \log_3(5x^{2}) log 3 x 2 + log 3 5 = log 3 ( 5 x 2 ) Why? Sum of logs → product.
Step 3. log 3 ( 5 x 2 ) − log 3 y = log 3 ( 5 x 2 y ) . \log_3(5x^{2}) - \log_3 y = \log_3\!\left(\dfrac{5x^{2}}{y}\right). log 3 ( 5 x 2 ) − log 3 y = log 3 ( y 5 x 2 ) . Why? Difference → quotient. ✅
Worked example Evaluate without a calculatorFind log 2 48 − log 2 3 \log_2 48 - \log_2 3 log 2 48 − log 2 3 Step 1. Quotient rule: = log 2 ( 48 3 ) = log 2 16 =\log_2\!\left(\tfrac{48}{3}\right)=\log_2 16 = log 2 ( 3 48 ) = log 2 16 Why? Difference collapses to a division we can simplify.
Step 2. 16 = 2 4 16=2^{4} 16 = 2 4 log 2 16 = 4 \log_2 16 = 4 log 2 16 = 4 Why? log b ( b k ) = k \log_b(b^k)=k log b ( b k ) = k
Worked example Solve an equationSolve 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. Product rule: log 5 ( x ( x − 4 ) ) = 1 \log_5\big(x(x-4)\big)=1 log 5 ( x ( x − 4 ) ) = 1 Step 2. Rewrite: x ( x − 4 ) = 5 1 = 5 x(x-4)=5^{1}=5 x ( x − 4 ) = 5 1 = 5 Why? log 5 A = 1 ⇔ A = 5 \log_5 A = 1 \Leftrightarrow A = 5 log 5 A = 1 ⇔ A = 5 Step 3. 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. Reject x = − 1 x=-1 x = − 1 log 5 ( x ) \log_5(x) log 5 ( x ) Answer x = 5 x=5 x = 5 Why? Domain: arguments must be > 0 >0 > 0
log ( M + N ) = log M + log N \log(M+N)=\log M+\log N log ( M + N ) = log M + log N Why it feels right: the product rule looks like a distributive law, so people distribute over + + + The fix: logs turn products into sums, not sums into sums. log b ( M N ) = log b M + log b N \log_b(MN)=\log_b M+\log_b N log b ( M N ) = log b M + log b N log b ( M + N ) \log_b(M+N) log b ( M + N ) cannot be simplified . Test: log 10 ( 1 + 1 ) = log 10 2 ≈ 0.30 \log_{10}(1+1)=\log_{10}2\approx0.30 log 10 ( 1 + 1 ) = log 10 2 ≈ 0.30 log 10 1 + log 10 1 = 0 \log_{10}1+\log_{10}1=0 log 10 1 + log 10 1 = 0
log M log N = log M − log N \dfrac{\log M}{\log N}=\log M-\log N log N log M = log M − log N Why it feels right: the quotient inside a log becomes a subtraction, so students subtract for a quotient of logs too.
The fix: subtraction only applies to log ( M / N ) \log(M/N) log ( M / N ) change-of-base result log M log N = log N M \dfrac{\log M}{\log N}=\log_N M log N log M = log N M
( log M ) 2 = 2 log M (\log M)^2 = 2\log M ( log M ) 2 = 2 log M Why it feels right: power rule brings exponents down — but only exponents on the argument M M M .
The fix: log ( M 2 ) = 2 log M \log(M^2)=2\log M log ( M 2 ) = 2 log M ( log M ) 2 = log M ⋅ log M (\log M)^2 = \log M \cdot \log M ( log M ) 2 = log M ⋅ log M whole log squared) — no simplification.
Common mistake Forgetting the domain when solvingWhy it feels right: you get two algebraic solutions and want both.
The fix: always check each answer keeps every logged argument > 0 >0 > 0
Recall Prove the product rule (cover the note, do it)Let p = log b M , q = log b N p=\log_b M,\ q=\log_b N p = log b M , q = log b N b p = M , b q = N b^p=M,\ b^q=N b p = M , b q = N M N = b p + q MN=b^{p+q} M N = b p + q log b ( M N ) = p + q = log b M + log b N . \log_b(MN)=p+q=\log_b M+\log_b N. log b ( M N ) = p + q = log b M + log b N .
Recall Which operation inside a log becomes subtraction?Division: log b ( M / N ) = log b M − log b N \log_b(M/N)=\log_b M-\log_b N log b ( M / N ) = log b M − log b N
Recall Explain like I'm 12 (Feynman)A log is a "how many times do I multiply?" counter. If you multiply two piles of stuff together, you just add how many multiplications each needed — so logs turn "times" into "plus". Dividing means you take some away, so logs turn "divide" into "minus". And "M 3 M^3 M 3 M M M 3 times the count for one M M M
Mnemonic Remember which law is which"Multiply, Add — Divide, Subtract — Power, Ping-out-front."
Or: logs are peace-makers — they turn the harder operation into the easier one below it (×→+, ÷→−, power→×).
What does log b x = y \log_b x = y log b x = y b y = x b^{y}=x b y = x (with
b > 0 , b ≠ 1 , x > 0 b>0,b\neq1,x>0 b > 0 , b = 1 , x > 0 ).
State the product rule for logs. log b ( M N ) = log b M + log b N \log_b(MN)=\log_b M+\log_b N log b ( M N ) = log b M + log b N .
State the quotient rule for logs. log b ( M / N ) = log b M − log b N \log_b(M/N)=\log_b M-\log_b N log b ( M / N ) = log b M − log b N .
State the power rule for logs. log b ( M k ) = k log b M \log_b(M^{k})=k\log_b M log b ( M k ) = k log b M for any real
k k k .
Key step that proves the product rule? Same-base multiplication adds exponents:
b p b q = b p + q b^{p}b^{q}=b^{p+q} b p b q = b p + q .
Key step that proves the power rule? Power of a power multiplies exponents:
( b p ) k = b p k (b^{p})^{k}=b^{pk} ( b p ) k = b p k .
Simplify log 2 48 − log 2 3 \log_2 48-\log_2 3 log 2 48 − log 2 3 log 2 16 = 4 \log_2 16 = 4 log 2 16 = 4 .
Is log ( M + N ) = log M + log N \log(M+N)=\log M+\log N log ( M + N ) = log M + log N No — logs turn products into sums, not sums into sums.
What is log M log N \dfrac{\log M}{\log N} log N log M log N M \log_N M log N M (change of base) — NOT
log M − log N \log M-\log N log M − log N .
Why reject some solutions when solving log equations? Every logged argument must be
> 0 >0 > 0 (domain restriction).
What is log b ( 1 / M ) \log_b(1/M) log b ( 1/ M ) − log b M -\log_b M − log b M (power rule with
k = − 1 k=-1 k = − 1 ).
Why must x > 0 x>0 x > 0 log b x \log_b x log b x A positive base to any real power is always positive, so non-positive
x x x has no real log.
Definition log_b x=y iff b^y=x
b^(log x)=x and log_b b^k=k
Restrictions b>0, b!=1, x>0
Product rule log MN = log M + log N
Quotient rule log M/N = log M - log N
Power rule log M^k = k log M
Intuition Hinglish mein samjhoDekho, log ka matlab bas ek sawaal hai: "log b x \log_b x log b x "base b b b x x x Yaani log actually ek exponent hi hota hai , chhupa hua. Yeh ek baat samajh gaye to poora chapter khul jaata hai, kyunki jitne bhi rules exponents (indices) ke hain, wahi log ke laws ban jaate hain.
Ab teen laws: jab tum do numbers ko multiply karte ho, to unke exponents add hote hain (b p ⋅ b q = b p + q b^p \cdot b^q = b^{p+q} b p ⋅ b q = b p + q log ( M N ) = log M + log N \log(MN)=\log M+\log N log ( M N ) = log M + log N Divide karo to exponents subtract hote hain — isliye log ( M / N ) = log M − log N \log(M/N)=\log M-\log N log ( M / N ) = log M − log N power lagao (M k M^k M k multiply ho jaata hai front pe — isliye log ( M k ) = k log M \log(M^k)=k\log M log ( M k ) = k log M p = log b M p=\log_b M p = log b M b p = M b^p=M b p = M
Exam me sabse zyada marks isi se aate hain: expressions ko ek single log me combine karna, ya tod-na. Jaise 2 log x + log 5 − log y = log 5 x 2 y 2\log x + \log 5 - \log y = \log\frac{5x^2}{y} 2 log x + log 5 − log y = log y 5 x 2 log ( M + N ) \log(M+N) log ( M + N ) log M + log N \log M+\log N log M + log N product ko sum banata hai, sum ko nahi. Aur log equation solve karne ke baad hamesha check karo ki har log ke andar ki value positive ho, warna woh solution reject karna padega.
