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Algebra - Logarithmic and Exponential Functions

Grade 12IGCSE

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

🔑Concepts

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Definition of an Exponential Function: A function of the form f(x)=axf(x) = a^x where a>0a > 0 and a≠1a \neq 1.

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Definition of a Logarithm: The inverse of exponentiation. If y=axy = a^x, then x=log⁡ayx = \log_a y.

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Natural Logarithms and Euler's Number: The base e≈2.718e \approx 2.718 is used for natural logarithms, denoted as ln⁡x=log⁡ex\ln x = \log_e x.

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Domain and Range: For y=log⁡axy = \log_a x, the domain is x>0x > 0 and the range is all real numbers.

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Graphs: Exponential functions axa^x grow/decay asymptotically to the x-axis, while logarithmic functions log⁡ax\log_a x are asymptotic to the y-axis.

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Change of Base: Transforming logarithms from one base to another to facilitate calculation using a calculator.

📐Formulae

log⁡a(xy)=log⁡ax+log⁡ay\log_a (xy) = \log_a x + \log_a y (Product Rule)

log⁡a(xy)=log⁡ax−log⁡ay\log_a (\frac{x}{y}) = \log_a x - \log_a y (Quotient Rule)

log⁡a(xk)=klog⁡ax\log_a (x^k) = k \log_a x (Power Rule)

log⁡aa=1\log_a a = 1 and log⁡a1=0\log_a 1 = 0

log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a} (Change of Base)

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

ax=b  ⟺  x=log⁡blog⁡aa^x = b \iff x = \frac{\log b}{\log a}

💡Examples

Problem 1:

Solve for xx: 52x−1=125^{2x-1} = 12. Give your answer to 3 decimal places.

Solution:

2x−1=log⁡512  ⟹  2x−1=ln⁡12ln⁡5  ⟹  2x=2.48491.6094+1  ⟹  2x=2.5439  ⟹  x≈1.2722x - 1 = \log_5 12 \implies 2x - 1 = \frac{\ln 12}{\ln 5} \implies 2x = \frac{2.4849}{1.6094} + 1 \implies 2x = 2.5439 \implies x \approx 1.272

Explanation:

To solve an exponential equation where the bases cannot be made the same, take the logarithm of both sides. Use the power rule to bring the exponent down, then isolate xx using algebraic manipulation.

Problem 2:

Simplify the expression: 2log⁡36−log⁡342\log_3 6 - \log_3 4.

Solution:

2log⁡36−log⁡34=log⁡3(62)−log⁡34=log⁡336−log⁡34=log⁡3(364)=log⁡39=22\log_3 6 - \log_3 4 = \log_3 (6^2) - \log_3 4 = \log_3 36 - \log_3 4 = \log_3 (\frac{36}{4}) = \log_3 9 = 2.

Explanation:

First, apply the Power Rule to move the coefficient into the exponent. Then, use the Quotient Rule for logarithms to combine the terms. Finally, evaluate the resulting logarithm.

Problem 3:

Solve the equation: log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2 (x - 2) = 3.

Solution:

log⁡2[x(x−2)]=3  ⟹  x2−2x=23  ⟹  x2−2x−8=0  ⟹  (x−4)(x+2)=0\log_2 [x(x - 2)] = 3 \implies x^2 - 2x = 2^3 \implies x^2 - 2x - 8 = 0 \implies (x - 4)(x + 2) = 0. Thus, x=4x = 4 or x=−2x = -2. However, xx must be >2> 2 for the logs to be defined, so x=4x = 4.

Explanation:

Use the Product Rule to combine the logarithms into a single term. Convert the logarithmic equation into its equivalent exponential form (ay=xa^y = x). Solve the resulting quadratic equation and always check for extraneous solutions (logs of negative numbers are undefined).