An exponential equation hides the unknown in the exponent , like 2 x = 7 2^x = 7 2 x = 7 x x x logarithm is the tool that "brings the exponent down to ground level" so you can solve for it.
WHY it works: logs and exponentials are inverse operations . Just as \sqrt{\ \ } log b \log_b log b b ( ) b^{(\ )} b ( )
Definition Exponential equationAn equation where the variable appears in an exponent , e.g. a ⋅ b f ( x ) = c a\cdot b^{f(x)} = c a ⋅ b f ( x ) = c logarithm .
WHAT we need: one key fact — the power law of logarithms.
Worked example Example 1 — the clean case:2 x = 7 2^x = 7 2 x = 7 Step 1: log ( 2 x ) = log 7 \log(2^x) = \log 7 log ( 2 x ) = log 7 Why? Apply log to both sides to expose the exponent.
Step 2: x log 2 = log 7 x\log 2 = \log 7 x log 2 = log 7 Why? Power law brings x x x Step 3: x = log 7 log 2 = 0.8451 0.3010 ≈ 2.807 x = \dfrac{\log 7}{\log 2} = \dfrac{0.8451}{0.3010} \approx 2.807 x = log 2 log 7 = 0.3010 0.8451 ≈ 2.807 Why? Divide by the constant log 2 \log 2 log 2 Check (Forecast-then-Verify): 2 2.807 ≈ 7.00 2^{2.807}\approx 7.00 2 2.807 ≈ 7.00 2 2 = 4 2^2=4 2 2 = 4 2 3 = 8 2^3=8 2 3 = 8
Worked example Example 2 — a coefficient in front:5 ⋅ 3 x = 60 5\cdot 3^x = 60 5 ⋅ 3 x = 60 First isolate the power. Divide by 5: 3 x = 12 3^x = 12 3 x = 12 Why? Log's power law only helps when the exponential term is alone; the 5 5 5 Step 1: log ( 3 x ) = log 12 \log(3^x) = \log 12 log ( 3 x ) = log 12 Step 2: x log 3 = log 12 x\log 3 = \log 12 x log 3 = log 12 Step 3: x = log 12 log 3 = 1.0792 0.4771 ≈ 2.262 x = \dfrac{\log 12}{\log 3} = \dfrac{1.0792}{0.4771}\approx 2.262 x = log 3 log 12 = 0.4771 1.0792 ≈ 2.262 3 2 = 9 < 12 < 27 = 3 3 3^2=9<12<27=3^3 3 2 = 9 < 12 < 27 = 3 3 2 < x < 3 2<x<3 2 < x < 3
Worked example Example 3 — variable in a bracket exponent:7 2 x − 1 = 40 7^{2x-1} = 40 7 2 x − 1 = 40 Step 1: log ( 7 2 x − 1 ) = log 40 \log(7^{2x-1}) = \log 40 log ( 7 2 x − 1 ) = log 40 Step 2: ( 2 x − 1 ) log 7 = log 40 (2x-1)\log 7 = \log 40 ( 2 x − 1 ) log 7 = log 40 Why? The whole exponent 2 x − 1 2x-1 2 x − 1 Step 3: 2 x − 1 = log 40 log 7 = 1.6021 0.8451 ≈ 1.8958 2x-1 = \dfrac{\log 40}{\log 7} = \dfrac{1.6021}{0.8451} \approx 1.8958 2 x − 1 = log 7 log 40 = 0.8451 1.6021 ≈ 1.8958 Step 4: 2 x = 2.8958 ⇒ x ≈ 1.448 2x = 2.8958 \Rightarrow x \approx 1.448 2 x = 2.8958 ⇒ x ≈ 1.448 Why? Now it's ordinary linear algebra.
Worked example Example 4 — same base both sides:2 x + 1 = 8 x 2^{x+1} = 8^{x} 2 x + 1 = 8 x Trick: write both as powers of 2. 8 = 2 3 8 = 2^3 8 = 2 3 8 x = 2 3 x 8^x = 2^{3x} 8 x = 2 3 x 2 x + 1 = 2 3 x 2^{x+1} = 2^{3x} 2 x + 1 = 2 3 x Why? If bases are equal, exponents must be equal.
Therefore x + 1 = 3 x ⇒ 1 = 2 x ⇒ x = 1 2 x+1 = 3x \Rightarrow 1 = 2x \Rightarrow x = \tfrac12 x + 1 = 3 x ⇒ 1 = 2 x ⇒ x = 2 1 Why this is smarter than logs here: matching bases avoids decimals entirely — use it whenever numbers are powers of a common base.
log ( 5 ⋅ 3 x ) = log 5 ⋅ log ( 3 x ) \log(5\cdot 3^x) = \log 5 \cdot \log(3^x) log ( 5 ⋅ 3 x ) = log 5 ⋅ log ( 3 x ) Why it feels right: log looks like it "distributes" over multiplication, like × \times × The truth: log ( m n ) = log m + log n \log(mn) = \log m + \log n log ( mn ) = log m + log n sums , not products. And you shouldn't log the 5 5 5 divide it off first.
Fix: isolate 3 x 3^x 3 x
x = log c log b x = \dfrac{\log c}{\log b} x = log b log c log ( c b ) \log\!\left(\dfrac{c}{b}\right) log ( b c ) Why it feels right: both have c c c b b b The truth: log c log b \dfrac{\log c}{\log b} log b log c ratio of two logs (change of base), while log c b = log c − log b \log\frac{c}{b} = \log c - \log b log b c = log c − log b difference . Totally different values.
Fix: log c log b = log b c \dfrac{\log c}{\log b}=\log_b c log b log c = log b c
Common mistake Forgetting the bracket:7 2 x − 1 = 40 ⇒ 2 x log 7 − 1 = log 40 7^{2x-1}=40 \Rightarrow 2x\log 7 - 1 = \log 40 7 2 x − 1 = 40 ⇒ 2 x log 7 − 1 = log 40 Why it feels right: you rushed and applied the power law to only part of the exponent.
The truth: the entire exponent ( 2 x − 1 ) (2x-1) ( 2 x − 1 ) log 7 \log 7 log 7 ( 2 x − 1 ) log 7 (2x-1)\log 7 ( 2 x − 1 ) log 7 Fix: always bracket the whole exponent before bringing it down.
Recall Feynman: explain to a 12-year-oldImagine a number is hiding at the top of a ladder (the exponent), and you can't reach it. A logarithm is like a special elevator that carries whatever is at the top all the way down to the floor, turning it into a normal number you can push around with easy arithmetic. So to solve "2 to the power of what equals 7?", you send in the log elevator, bring the what down, and then just divide.
Mnemonic Remember the recipe"ILDS" — I solate the power, L og both sides, D rop the exponent (power law), S olve.
Say it: "I Love Dropping Stuff (down from the exponent)."
Only one law does the heavy lifting: log ( m k ) = k log m \log(m^k)=k\log m log ( m k ) = k log m
Isolate the exponential first , then log.Result is always x = log c log b x=\dfrac{\log c}{\log b} x = log b log c ratio , never a difference.
If both sides share a base, match exponents instead (cleaner, exact).
What operation "undoes" an exponential, letting you solve for a trapped exponent? Taking a logarithm of both sides (log and exponential are inverse operations).
State the power law of logarithms. log b ( m k ) = k log b ( m ) \log_b(m^k) = k\log_b(m) log b ( m k ) = k log b ( m ) — the exponent becomes a multiplier.
Derive why log b ( m k ) = k log b m \log_b(m^k)=k\log_b m log b ( m k ) = k log b m Let
y = log b m y=\log_b m y = log b m so
b y = m b^y=m b y = m ; raise to
k k k :
b y k = m k b^{yk}=m^k b y k = m k ; in log form
log b ( m k ) = y k = k log b m \log_b(m^k)=yk=k\log_b m log b ( m k ) = y k = k log b m .
Solve 2 x = 7 2^x=7 2 x = 7 x log 2 = log 7 x\log2=\log7 x log 2 = log 7 , so
x = log 7 / log 2 ≈ 2.807 x=\log7/\log2\approx2.807 x = log 7/ log 2 ≈ 2.807 .
For 5 ⋅ 3 x = 60 5\cdot3^x=60 5 ⋅ 3 x = 60 Divide by 5 to isolate the power:
3 x = 12 3^x=12 3 x = 12 .
Why can't you write x = log ( c / b ) x=\log(c/b) x = log ( c / b ) log c / log b \log c/\log b log c / log b log c / log b \log c/\log b log c / log b is a ratio (=
log b c \log_b c log b c );
log ( c / b ) = log c − log b \log(c/b)=\log c-\log b log ( c / b ) = log c − log b is a difference — different values.
Solve 2 x + 1 = 8 x 2^{x+1}=8^x 2 x + 1 = 8 x 8 = 2 3 8=2^3 8 = 2 3 , so
x + 1 = 3 x x+1=3x x + 1 = 3 x , giving
x = 1 2 x=\tfrac12 x = 2 1 .
In 7 2 x − 1 = 40 7^{2x-1}=40 7 2 x − 1 = 40 The whole exponent
( 2 x − 1 ) (2x-1) ( 2 x − 1 ) as a bracket:
( 2 x − 1 ) log 7 = log 40 (2x-1)\log7=\log40 ( 2 x − 1 ) log 7 = log 40 .
The general solution of b x = c b^x=c b x = c x = log c log b = log b c x=\dfrac{\log c}{\log b}=\log_b c x = log b log c = log b c (any log base works).
Power law log_b m^k = k log_b m
Definition of log b^y = m
Intuition Hinglish mein samjhoDekho, exponential equation ka matlab hai jaha variable "power" me chhupa hota hai, jaise 2 x = 7 2^x = 7 2 x = 7 x x x logarithm — kyunki log aur exponential ek dusre ke inverse hain. Jaise square root, square ko undo karta hai, waise hi log b \log_b log b b ( ) b^{(\ )} b ( )
Sabse important rule ek hi hai: power law , yaani log ( m k ) = k log m \log(m^k)=k\log m log ( m k ) = k log m ILDS : pehle power ko Isolate karo (agar aage koi number multiply ho raha hai to usse divide karo), phir dono side Log lo, phir exponent ko Drop karo (power law), aur last me Solve karo. Answer hamesha aata hai x = log c log b x=\dfrac{\log c}{\log b} x = log b log c
Ek common galti: log ko multiplication pe "distribute" mat karo — log ( m n ) = log m + log n \log(mn)=\log m+\log n log ( mn ) = log m + log n log c log b \dfrac{\log c}{\log b} log b log c log ( c / b ) \log(c/b) log ( c / b ) 8 = 2 3 8=2^3 8 = 2 3
