1.2.5. frac{log_{a} 16 - log_{b} 4}{log_{a} 4 - log_{b} 2} 1.3. Determine the value of a if f(x) = x^3 + ax^2 - x + 5 is divided by x - 2 and gives a remainder of 23. Question 2 2.1. Solve for x: 2.1.1. log_{2} x - 3 log_{x} 2 = 2 2.1.2. frac{25^{-2x+1} imes 5^{frac{x}{2}}}{125^{x-1}} = 1 2.1.2. frac{2^{2x-1} - 2^{2x}}{4^x} = sqrt{2^{-3x}} - 1 2.2. Solve for x by completing the square: -4x = -4x^2 + 1 Question 3 3.1. Make R the subject of the following formula: frac{RMN}{RE + MA} = 1 3.2. When 1 is subtracted from the numerator and denominator of a fraction, the fraction becomes frac{1}{2}. When 2 is added to the numerator and the denominator of the same fraction, the fraction becomes frac{1}{5}. Determine the fraction.
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2.5. Using the properties of logarithms, we can simplify the expression as follows: \( \frac{\log _{a} 16-\log _{b} 4}{\log _{a} 4-\log _{b} 2} = \frac{\log _{a} (16/4)}{\log _{a} (4/2)} = \frac{\log _{a} 4}{\log _{a} 2} = 2 \) Show more…
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Task 5. Using the Laws of Logarithms to Solve Logarithmic Equations. We will use the Laws of Logarithms to simplify the equation as far as possible and then use the definition to solve for the unknown argument. Here is an example: Solve the equation log2(x) - log2(x - 3) = 1. Simplify the equation: log2(x / (x - 3)) = 1 Use the definition, the inverse function, to solve for the argument: x / (x - 3) = 2^1. Continue to solve the algebraic equation: x = 2x - 6 x = 6 Check the solution to see that it is in the domain: log2(6) - log2(6 - 3) = 1 log2(6) - log2(3) = 1 You can proceed to check this further using a calculator or a hand calculation. But it should work since we can see that the terms log2(6) and log2(3) are defined, that is we know that both 6 and 3 are in the domain of the logarithm. Solve the following equations. Be sure to check the potential solution(s) in the original equation, to see whether it (they) are in the domain: (a) log2(x^2 - x - 2) = 2 (b) 2log(x) - log(2) - log(3x - 4) = 0 (c) log(x) + log(x - 3) = 1
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Fill in the reason for each step in the following two solutions. Solve: $\log _{3}(x-1)^{2}=2$ $$ \begin{aligned} &\begin{array}{ll} {\text { Solution } A} & {\text { Solution } B} \\ {\log _{3}(x-1)^{2}=2} & {\log _{3}(x-1)^{2}=2} \end{array}\\ &(x-1)^{2}=3^{2}=9 \quad 2 \log _{3}(x-1)=2\\ &(x-1)=\pm 3 \quad \log _{3}(x-1)=1\\ &x-1=-3 \text { or } x-1=3 \quad x-1=3^{1}=3\\ &x=-2 \text { or } x=4\\ &\pi x=4 \end{aligned} $$ Both solutions given in Solution A check. Explain what caused the solution $x=-2$ to be lost in Solution B.
Exponential and Logarithmic Functions
Logarithmic and Exponential Equations
Solve the equation $2^{x+1}=3^{2 x-5}$ correct to 2 decimal places. Taking logarithms to base 10 of both sides gives: $$ \log _{10} 2^{x+1}=\log _{10} 3^{2 x-5} $$ i.e. $\quad(x+1) \log _{10} 2=(2 x-5) \log _{10} 3$ $$ \begin{gathered} x \log _{10} 2+\log _{10} 2=2 x \log _{10} 3-5 \log _{10} 3 \\ x(0.3010)+(0.3010)=2 x(0.4771)-5(0.4771) \end{gathered} $$ i.e. $\quad 0.3010 x+0.3010=0.9542 x-2.3855$ Hence $$ \begin{aligned} 2.3855+0.3010 &=0.9542 x-0.3010 x \\ 2.6865 &=0.6532 x \end{aligned} $$ from which $x=\frac{2.6865}{0.6532}=4.11$, correct to 2 decimal places.
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