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Number and Algebra - Exponents and Logarithms

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Laws of Exponents: These fundamental rules simplify operations with powers. For multiplication, aman=am+na^m \cdot a^n = a^{m+n}, and for division, aman=amn\frac{a^m}{a^n} = a^{m-n}. When a power is raised to another power, (am)n=amn(a^m)^n = a^{mn}. Visually, an exponential growth function f(x)=axf(x) = a^x where a>1a > 1 appears as a curve that rises increasingly steeply from left to right, crossing the y-axis at (0,1)(0, 1) and approaching the x-axis as a horizontal asymptote (y=0y=0) for negative values of xx.

Logarithmic and Exponential Relationship: A logarithm is the inverse of an exponentiation. The statement x=ayx = a^y is equivalent to y=logaxy = \log_a x. Graphically, the logarithmic function y=logaxy = \log_a x is a reflection of the exponential function y=axy = a^x across the line y=xy = x. While exponential graphs have horizontal asymptotes, logarithmic graphs have a vertical asymptote at x=0x = 0, meaning the graph approaches but never touches the y-axis.

Laws of Logarithms: These rules allow for the expansion and condensation of logarithmic expressions. The Product Law states loga(xy)=logax+logay\log_a (xy) = \log_a x + \log_a y, the Quotient Law states loga(xy)=logaxlogay\log_a (\frac{x}{y}) = \log_a x - \log_a y, and the Power Law states loga(xk)=klogax\log_a (x^k) = k \log_a x. These laws are essential for solving equations where the unknown variable is located in the exponent.

The Natural Logarithm and 'e': The irrational number e2.718e \approx 2.718 is the base for the natural logarithm, written as lnx\ln x (which means logex\log_e x). The function f(x)=exf(x) = e^x is unique because its gradient at any point is equal to its y-value. In real-world contexts, ee is frequently used to model continuous growth or decay, such as population dynamics or radioactive half-life.

Change of Base Formula: To calculate or simplify logarithms with bases not available on a standard calculator, we use the formula logab=logcblogca\log_a b = \frac{\log_c b}{\log_c a}. This allows any logarithm to be converted into a base that is easier to work with, typically base 10 (common log) or base ee (natural log).

Negative and Rational Exponents: A negative exponent represents a reciprocal: an=1ana^{-n} = \frac{1}{a^n}. A rational exponent represents a root: a1n=ana^{\frac{1}{n}} = \sqrt[n]{a} and amn=amna^{\frac{m}{n}} = \sqrt[n]{a^m}. Visually, as the root nn increases for x1nx^{\frac{1}{n}}, the curve becomes flatter for x>1x > 1 while still passing through the anchor point (1,1)(1, 1).

Solving Exponential Equations: When solving for xx in equations like ax=ba^x = b, we apply logarithms to both sides. This transforms the exponent into a coefficient using the Power Law: xloga=logbx \log a = \log b. If the bases on both sides of an equation can be written as powers of the same number (e.g., 2x=82^x = 8), we equate the exponents directly (2x=23    x=32^x = 2^3 \implies x = 3).

📐Formulae

aman=am+na^m \cdot a^n = a^{m+n}

aman=amn\frac{a^m}{a^n} = a^{m-n}

(am)n=amn(a^m)^n = a^{mn}

amn=amna^{\frac{m}{n}} = \sqrt[n]{a^m}

an=1ana^{-n} = \frac{1}{a^n}

a0=1,a0a^0 = 1, a \neq 0

logax=y    ay=x\log_a x = y \iff a^y = x

loga(xy)=logax+logay\log_a (xy) = \log_a x + \log_a y

loga(xy)=logaxlogay\log_a (\frac{x}{y}) = \log_a x - \log_a y

loga(xk)=klogax\log_a (x^k) = k \log_a x

logaa=1\log_a a = 1 and loga1=0\log_a 1 = 0

logab=lnblna\log_a b = \frac{\ln b}{\ln a}

elnx=xe^{\ln x} = x and ln(ex)=x\ln(e^x) = x

💡Examples

Problem 1:

Solve for xx: 32x1=53^{2x-1} = 5

Solution:

  1. Take the natural logarithm (ln\ln) of both sides: ln(32x1)=ln(5)\ln(3^{2x-1}) = \ln(5).
  2. Use the Power Law to move the exponent: (2x1)ln(3)=ln(5)(2x-1) \ln(3) = \ln(5).
  3. Divide by ln(3)\ln(3): 2x1=ln(5)ln(3)2x - 1 = \frac{\ln(5)}{\ln(3)}.
  4. Add 1 to both sides: 2x=ln(5)ln(3)+12x = \frac{\ln(5)}{\ln(3)} + 1.
  5. Divide by 2: x=12(ln(5)ln(3)+1)x = \frac{1}{2} (\frac{\ln(5)}{\ln(3)} + 1).
  6. Calculate the decimal value: x1.23x \approx 1.23 (to 3 significant figures).

Explanation:

To solve an exponential equation where the bases cannot be easily matched, apply a logarithm to both sides to 'bring down' the exponent, then use algebraic manipulation to isolate the variable.

Problem 2:

Solve for xx: log2(x)+log2(x2)=3\log_2(x) + \log_2(x-2) = 3

Solution:

  1. Use the Product Law to combine the logarithms: log2(x(x2))=3\log_2(x(x-2)) = 3.
  2. Convert the logarithmic equation into exponential form: x(x2)=23x(x-2) = 2^3.
  3. Expand and simplify: x22x=8x^2 - 2x = 8.
  4. Set the quadratic equation to zero: x22x8=0x^2 - 2x - 8 = 0.
  5. Factor the quadratic: (x4)(x+2)=0(x-4)(x+2) = 0.
  6. Identify potential solutions: x=4x = 4 or x=2x = -2.
  7. Check for extraneous solutions: The argument of a logarithm must be positive (x>0x > 0 and x2>0x-2 > 0). Therefore, x=2x = -2 is rejected. The final solution is x=4x = 4.

Explanation:

This problem requires condensing multiple log terms into one using log laws, converting the resulting expression into an exponential format to create a quadratic equation, and finally checking the validity of solutions against the domain of the original log functions.