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Number and Algebra - Exponents and Logarithms

Grade 11IB

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

🔑Concepts

Laws of Indices: These rules define how to manipulate expressions with exponents. When multiplying terms with the same base, add the exponents (am×an=am+na^m \times a^n = a^{m+n}); when dividing, subtract them (aman=amn\frac{a^m}{a^n} = a^{m-n}). Raising a power to another power involves multiplying the exponents ((am)n=amn(a^m)^n = a^{mn}). Visually, an exponent represents repeated multiplication, and these laws simplify the counting of those repetitions.

Rational and Negative Exponents: A negative exponent represents the reciprocal of the base raised to the positive power (an=1ana^{-n} = \frac{1}{a^n}). Fractional exponents represent roots: the denominator indicates the root degree and the numerator indicates the power (amn=amna^{\frac{m}{n}} = \sqrt[n]{a^m}). Visually, a12a^{\frac{1}{2}} represents the side length of a square with area aa.

Logarithm Definition and Inverse Relationship: A logarithm is the inverse of an exponential function. The statement loga(x)=b\log_a(x) = b is equivalent to ab=xa^b = x. Visually, if you look at the graph of y=axy = a^x, its reflection over the diagonal line y=xy = x produces the graph of y=loga(x)y = \log_a(x). This means the domain of the exponential (all real numbers) becomes the range of the logarithm, and the range of the exponential (y>0y > 0) becomes the domain of the logarithm (x>0x > 0).

Properties of Logarithms: Logarithms follow specific rules derived from index laws: the Product Rule (loga(xy)=logax+logay\log_a(xy) = \log_a x + \log_a y), the Quotient Rule (loga(xy)=logaxlogay\log_a(\frac{x}{y}) = \log_a x - \log_a y), and the Power Rule (loga(xk)=klogax\log_a(x^k) = k \log_a x). These properties allow complex multiplicative relationships to be converted into simpler additive ones.

Natural Logarithms and Euler's Number: The constant e2.71828e \approx 2.71828 is a transcendental number that serves as the base for the natural logarithm, denoted as ln(x)=loge(x)\ln(x) = \log_e(x). In calculus and growth modeling, exe^x is unique because its rate of change (slope) at any point is equal to the value of the function itself. Visually, the graph of y=exy = e^x passes through (0,1)(0, 1) with a gradient of 11.

Solving Exponential Equations: When variables are in the exponent, such as ax=ba^x = b, we apply logarithms to both sides to 'bring down' the exponent using the power rule. If the bases can be made equal, we can simply equate the exponents. Visually, solving ax=ba^x = b is finding the xx-coordinate where the exponential curve y=axy = a^x intersects the horizontal line y=by = b.

Logarithmic Graphs and Asymptotes: The graph of f(x)=loga(x)f(x) = \log_a(x) always passes through the point (1,0)(1, 0) because any base raised to the power 00 is 11. It has a vertical asymptote at x=0x = 0 (the yy-axis), meaning the function is undefined for non-positive values of xx. As xx increases, the graph increases but the slope becomes flatter and flatter, representing very slow growth.

Change of Base Formula: To calculate or simplify logarithms with bases not available on a standard calculator, the change of base formula is used: loga(b)=logc(b)logc(a)\log_a(b) = \frac{\log_c(b)}{\log_c(a)}. This allows any logarithm to be expressed in terms of common logarithms (base 1010) or natural logarithms (base ee).

📐Formulae

am×an=am+na^m \times 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}

ax=b    x=loga(b)a^x = b \iff x = \log_a(b)

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

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

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

loga(b)=ln(b)ln(a)=log10(b)log10(a)\log_a(b) = \frac{\ln(b)}{\ln(a)} = \frac{\log_{10}(b)}{\log_{10}(a)}

loga(1)=0\log_a(1) = 0 and loga(a)=1\log_a(a) = 1

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

💡Examples

Problem 1:

Solve the equation 52x1=125^{2x-1} = 12 for xx. Give your answer correct to 3 significant figures.

Solution:

Step 1: Take the natural logarithm of both sides: ln(52x1)=ln(12)\ln(5^{2x-1}) = \ln(12) Step 2: Use the power rule for logarithms to move the exponent: (2x1)ln(5)=ln(12)(2x - 1) \ln(5) = \ln(12) Step 3: Isolate the term with xx by dividing by ln(5)\ln(5): 2x1=ln(12)ln(5)2x - 1 = \frac{\ln(12)}{\ln(5)} Step 4: Solve for xx: 2x=ln(12)ln(5)+12x = \frac{\ln(12)}{\ln(5)} + 1 x=12(ln(12)ln(5)+1)x = \frac{1}{2} \left( \frac{\ln(12)}{\ln(5)} + 1 \right) Step 5: Calculate the numerical value: x12(1.54395+1)1.27197x \approx \frac{1}{2} (1.54395 + 1) \approx 1.27197 Rounding to 3 sig figs: x1.27x \approx 1.27.

Explanation:

This problem uses the logarithmic power rule to solve an exponential equation where the bases cannot be easily matched. Taking the log of both sides allows us to treat the exponent as a linear coefficient.

Problem 2:

Simplify the expression 2log3(6)log3(4)+log3(1)2\log_3(6) - \log_3(4) + \log_3(1).

Solution:

Step 1: Simplify the third term. Since loga(1)=0\log_a(1) = 0 for any base aa: log3(1)=0\log_3(1) = 0 Step 2: Use the power rule on the first term: 2log3(6)=log3(62)=log3(36)2\log_3(6) = \log_3(6^2) = \log_3(36) Step 3: Combine the remaining terms using the quotient rule: log3(36)log3(4)=log3(364)\log_3(36) - \log_3(4) = \log_3\left(\frac{36}{4}\right) Step 4: Evaluate the fraction and the resulting log: log3(9)=log3(32)=2\log_3(9) = \log_3(3^2) = 2 Final Answer: 22.

Explanation:

This solution demonstrates the use of the Power Rule and Quotient Rule to combine multiple logarithmic terms into a single value. It also utilizes the fundamental property that loga(1)=0\log_a(1)=0.