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Grade 11 / High 2 (age 16-17)

high school Exponents & Logarithms

Exponents & Logarithms

A logarithm asks the reverse of an exponent. log₂8 means how many times you multiply 2 to reach 8, and the answer is 3, so the two are sides of one coin. By turning multiplication into addition, it tames huge numbers and spans wide ranges like loudness and earthquakes. Drag the base and the curve climbs, and a log scale spaces multiplication evenly.

Exponent Intuition — Repeated Multiplication

When you know the base and the power but not the exponent, the missing number needs a name. That name is log_a N: it is the x that solves a^x = N. Because a product of the same base turns into a sum of exponents, a log turns a long chain of multiplications into addition. The base must satisfy a > 0 and a ≠ 1, and the argument must satisfy N > 0. The slider only allows bases from 1.2 to 4, so the sketch shows growth, not decay. The case 0 < a < 1 is written in the text; it is outside the drawn range.

The upper sketch shows a gold curve y = a^x, with x ticks from 0 to 5. A dashed line sits at height 1, and the title prints the chosen base to one decimal place in y = a^x. When you change base a from 1.2 to 4, the vertical scale is reset to height a^5, so a larger base stays low on the left and then climbs steeply on the right. Below, the same five values a^0, a^1, a^2, a^3, a^4 are marked on two bars. The upper bar is a log scale, so the gaps are even; the lower bar is linear, so larger numbers bunch toward one end. The curve has no extra marker point. Each new base redraws the curve and the ticks in place.

2
💡 Exponentiation is repeated multiplication
①2^3 = 2×2×2 = 8: 'multiply 2 three times'
②Larger base → explosive growth
③Base > 1 grows, 0 < base < 1 shrinks
④This sketch uses a ≥ 1.2. Decay for 0 < a < 1 is the sentence in

Logarithm Intuition — The Inverse Question

If you swap the two places in log_a N, you are asking a different question. The first number is the base and the second is the argument; flip them and you have inverted the power. A base of 1, a negative base, or an argument that is 0 or less is not defined, so a familiar-looking symbol is not enough to compute. The rule that turns a log of a product into a sum of logs holds only when the argument is a product. Using that rule on a sum breaks the definition. Dropping the base and reading only the argument also makes the same letters stand for another value.

💡 Logs ask 'how many times do I multiply?'
①log_2(8) = 3 → 'how many times multiply 2 to get 8?' → 3 times
②Multiplications appear evenly spaced on a log scale
③Earthquake magnitude, decibels, pH — all log scales

Laws of Exponents & Logs

Exponent Laws
am × an = am+n, am ÷ an = am-n
Same base product → add exponents
Definition of Log
ax = N ⟺ x = loga N
How many times must a be raised to give N
Log Laws
loga MN = loga M + loga N
Log of a product = sum of logs
Change of Base
loga b = logc blogc a
Any base c works for conversion

Exponent ↔ Log Relationship

🔗 Exponents and logs are mirror images
①Exponent: base + exponent → value (2^3 = 8)
②Log: base + value → exponent (log_2 8 = 3)
③Knowing one immediately gives the other
④Graphs are reflections across y = x
⑤Common vs natural logs are distinguished later in this chapter
Core Identities
alog_a N = N, loga ax = x
Exponent and log are mutual inverses

Wrap-up

Core
loga N = x ⟺ ax = N
Logarithm is the inverse of exponentiation
🎯 Exam Points
①Definition: a^x = N ⟺ log_a N = x
②Conditions: a > 0, a ≠ 1, N > 0
③Log laws: product→sum, quotient→difference, power→coefficient
④Change of base: log_a b = log_c b / log_c a
⑤Distinguish common log (log₁₀) from natural log (ln)

Worked Examples & Past Exam

Example 1
Evaluate log2 8 + log2 4.
1
A sum of same-base logs combines into the log of a product.
log2 8 + log2 4 = log2 (8 × 4) = log2 32
2
Since 32 = 25, the log equals the exponent 5.
log2 32 = log2 25 = 5
5
Combine same-base logs into a product, then write it as a power.
Example 2
If log2 3 = a, express log2 24 in terms of a.
1
Factor 24 = 23 × 3.
log2 24 = log2 (23 × 3) = log2 23 + log2 3
2
log2 23 = 3 and log2 3 = a.
= 3 + a
a + 3
Split the argument into a power of the base times the remaining factor.
exam-style
If log3 5 = a and log3 4 = b, express log3 80 in terms of a and b. (Note: 80 = 16 × 5)
a + b
a + 2b
2a + b
ab
2a + 2b
② a + 2b
1
Factor 80 = 42 × 5.
log3 80 = log3 (42 × 5) = log3 42 + log3 5
2
log3 42 = 2 log3 4 = 2b and log3 5 = a.
= 2b + a = a + 2b
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Irrational Functions
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Exponential Function
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