Understand logarithms as inverse of exponentials. Math 30-1 Alberta curriculum.
Lesson 3.3 of Exponents & Logarithms in Math 30-1 — Alberta curriculum lessons.
In previous lessons you learned what exponential functions look like and how to transform them. Now comes the most important skill: solving for x when x is in the exponent. The entire strategy for this lesson comes down to one powerful idea:
If you can write both sides of an equation with the same base, the exponents must be equal and you can solve for x algebraically.
Before we get there, you need to have your exponent laws sharp. Every technique in this lesson depends on them.
For all of these, the letters u and v represent exponents and the letters a, b, c, d represent bases.
Product Law. a^u · a^v = a^(u+v)When multiplying powers with the same base, add the exponents.
Example: 3^2 · 3^5 = 3^(2+5) = 3^7
Quotient Law. (a^u)/(a^v) = a^(u-v)When dividing powers with the same base, subtract the exponents.
Example: (5^8)/(5^3) = 5^(8-3) = 5^5
Power of a Power Law. (b^u)^v = b^(uv)When raising a power to another power, multiply the exponents.
Example: (2^3)^4 = 2^(3 × 4) = 2^(12)
Power of a Product Law. (c^a · d^b)^u = c^(au) · d^(bu)Distribute the outer exponent to every factor inside the brackets.
Example: (2^3 · 5^2)^4 = 2^(12) · 5^8
Power of a Quotient Law. ((c^a)/(d^b))^u = c^(au)d^(bu), d ≠ 0Same idea as the Power of a Product Law, but applied to a fraction — distribute the outer exponent to both numerator and denominator.
Negative Exponent Law. b^(-u) = (1)/(b^u) and 1b^(-u) = b^u, b ≠ 0A negative exponent means "take the reciprocal."
Example: 2^(-3) = (1)/(2^3) = (1)/(8)
Zero Exponent Law. b^0 = 1, b ≠ 0Anything to the power of zero is 1.
Example: 7^0 = 1
Rational Exponent Law. b^(u/v) = √(b^u), v ≠ 0A fractional exponent means a root combined with a power. The denominator is the index of the root; the numerator is the power.
Example: 8^(2/3) = (√(8))^2 = 2^2 = 4
Step 1: Recognize that 49 is a power of 7.
49 = 7^2 (since 7 × 7 = 49)
Step 2: Rewrite 49^(2x) using this fact:
49^(2x) = (7^2)^(2x)
Step 3: Apply the Power of a Power Law — multiply the exponents:
(7^2)^(2x) = 7^(2 × 2x) = 7^(4x)
Step 4: Now deal with the 7 out front. Since 7 = 7^1:
7^1 · 7^(4x)
Step 5: Apply the Product Law — add the exponents:
7^1 · 7^(4x) = 7^(1+4x) = 7^(4x+1)
Answer. 7 · 49^(2x) = 7^(4x+1)
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