7 Solve each of the following equations.
a $\log _{3} x+\log _{3} 5=\log _{3}(2 x+3)$
b $\log _{9} x+\log _{9} 10=\frac{3}{2}$
c $\log _{4} x-\log _{4}(x-1)=\log _{4} 3+\frac{1}{2}$
d $\log _{5} 5 x-\log _{5}(x+2)=\log _{5}(x+6)-\log _{5} x$
e $2 \log _{6} x=\log _{6}(2 x-5)+\log _{6} 5$
f $\log _{7} 4 x=\log _{7} \frac{1}{x-6}+1$
8 Solve each pair of simultaneous equations.
a $\log _{x} y=2$ $x y=27$
b $\log _{5} x-2 \log _{5} y=\log _{5} 2$ $x+y^{2}=12$
c $\log _{2} x=3-2 \log _{2} y$
d $\log _{y} x=\frac{3}{2}$
$\log _{y} 32=-\frac{5}{2}$
$x^{\frac{1}{3}}+3 y^{\frac{1}{2}}=20$
e $\log _{a} x+\log _{a} 3=\frac{1}{2} \log _{a} y$
f $\log _{10} y+2 \log _{10} x=3$ $3 x+y=20$
$\log _{2} y-\log _{2} x=3$

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Accepted Answer

∻Solution by Steps∻ ### Question 7a ‖ step 1 ⋮ Simplify the given equation using the properties of logarithms: $ \log_{3} x + \log_{3} 5 = \log_{3} (2x + 3) $ ‖ ‖ step 2 ⋮ Combine the logarithms on the left side: $ \log_{3} (5x) = \log_{3} (2x + 3) $ ‖ ‖ step 3 ⋮ Since the bases and the arguments are equal, set the arguments equal to each other: $ 5x = 2x + 3 $ ‖ ‖ step 4 ⋮ Solve for $ x $ by isolating it on one side: $ 5x - 2x = 3 \implies 3x = 3 \implies x = 1 $ ‖ ### Question 7b ‖ step 1 ⋮ Simplify the given equation using the properties of logarithms: $ \log_{9} x + \log_{9} 10 = \frac{3}{2} $ ‖ ‖ step 2 ⋮ Combine the logarithms on the left side: $ \log_{9} (10x) = \frac{3}{2} $ ‖ ‖ step 3 ⋮ Rewrite the equation in exponential form: $ 10x = 9^{3/2} $ ‖ ‖ step 4 ⋮ Simplify the right side: $ 9^{3/2} = (3^2)^{3/2} = 3^3 = 27 $ ‖ ‖ step 5 ⋮ Solve for $ x $ by isolating it on one side: $ 10x = 27 \implies x = \frac{27}{10} = 2.7 $ ‖ ### Question 7c ‖ step 1 ⋮ Simplify the given equation using the properties of logarithms: $ \log_{4} x - \log_{4} (x - 1) = \log_{4} 3 + \frac{1}{2} $ ‖ ‖ step 2 ⋮ Combine the logarithms on the left side: $ \log_{4} \left( \frac{x}{x - 1} \right) = \log_{4} (3 \cdot \sqrt{4}) $ ‖ ‖ step 3 ⋮ Simplify the right side: $ \log_{4} (3 \cdot 2) = \log_{4} 6 $ ‖ ‖ step 4 ⋮ Since the bases and the arguments are equal, set the arguments equal to each other: $ \frac{x}{x - 1} = 6 $ ‖ ‖ step 5 ⋮ Solve for $ x $ by isolating it on one side: $ x = 6(x - 1) \implies x = 6x - 6 \implies 5x = 6 \implies x = \frac{6}{5} = 1.2 $ ‖ ### Question 7d ‖ step 1 ⋮ Simplify the given equation using the properties of logarithms: $ \log_{5} 5x - \log_{5} (x + 2) = \log_{5} (x + 6) - \log_{5} x $ ‖ ‖ step 2 ⋮ Combine the logarithms on both sides: $ \log_{5} \left( \frac{5x}{x + 2} \right) = \log_{5} \left( \frac{x + 6}{x} \right) $ ‖ ‖ step 3 ⋮ Since the bases and the arguments are equal, set the arguments equal to each other: $ \frac{5x}{x + 2} = \frac{x + 6}{x} $ ‖ ‖ step 4 ⋮ Cross-multiply to solve for $ x $: $ 5x^2 = (x + 2)(x + 6) \implies 5x^2 = x^2 + 8x + 12 $ ‖ ‖ step 5 ⋮ Simplify and solve the quadratic equation: $ 4x^2 - 8x - 12 = 0 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0 $ ‖ ‖ step 6 ⋮ Solve for $ x $: $ x = 3 $ or $ x = -1 $ (since $ x $ must be positive, $ x = 3 $) ‖ ### Question 7e ‖ step 1 ⋮ Simplify the given equation using the properties of logarithms: $ 2 \log_{6} x = \log_{6} (2x - 5) + \log_{6} 5 $ ‖ ‖ step 2 ⋮ Combine the logarithms on the right side: $ 2 \log_{6} x = \log_{6} (5(2x - 5)) $ ‖ ‖ step 3 ⋮ Simplify the right side: $ 2 \log_{6} x = \log_{6} (10x - 25) $ ‖ ‖ step 4 ⋮ Rewrite the left side: $ \log_{6} (x^2) = \log_{6} (10x - 25) $ ‖ ‖ step 5 ⋮ Since the bases and the arguments are equal, set the arguments equal to each other: $ x^2 = 10x - 25 $ ‖ ‖ step 6 ⋮ Solve the quadratic equation: $ x^2 - 10x + 25 = 0 \implies (x - 5)^2 = 0 \implies x = 5 $ ‖ ### Question 7f ‖ step 1 ⋮ Simplify the given equation using the properties of logarithms: $ \log_{7} 4x = \log_{7} \left( \frac{1}{x - 6} \right) + 1 $ ‖ ‖ step 2 ⋮ Rewrite the right side: $ \log_{7} 4x = \log_{7} \left( \frac{1}{x - 6} \right) + \log_{7} 7 $ ‖ ‖ step 3 ⋮ Combine the logarithms on the right side: $ \log_{7} 4x = \log_{7} \left( \frac{7}{x - 6} \right) $ ‖ ‖ step 4 ⋮ Since the bases and the arguments are equal, set the arguments equal to each other: $ 4x = \frac{7}{x - 6} $ ‖ ‖ step 5 ⋮ Cross-multiply to solve for $ x $: $ 4x(x - 6) = 7 \implies 4x^2 - 24x - 7 = 0 $ ‖ ‖ step 6 ⋮ Solve the quadratic equation using the quadratic formula: $ x = \frac{24 \pm \sqrt{24^2 + 4 \cdot 4 \cdot 7}}{2 \cdot 4} = \frac{24 \pm \sqrt{784}}{8} = \frac{24 \pm 28}{8} $ ‖ ‖ step 7 ⋮ Simplify the solutions: $ x = \frac{52}{8} = 6.5 $ or $ x = \frac{-4}{8} = -0.5 $ (since $ x $ must be positive, $ x = 6.5 $) ‖ ∻Answer∻ ⚹ 7a: $ x = 1 $ ⚹ ⚹ 7b: $ x = 2.7 $ ⚹ ⚹ 7c: $ x = 1.2 $ ⚹ ⚹ 7d: $ x = 3 $ ⚹ ⚹ 7e: $ x = 5 $ ⚹ ⚹ 7f: $ x = 6.5 $ ⚹ ∻Key Concept∻ ⚹Logarithmic Equations⚹ ∻Explanation∻ ⚹To solve logarithmic equations, use properties of logarithms to combine or simplify terms, then convert to exponential form if necessary and solve for the variable.⚹◊What is the solution to the equation $\log _{3} x+ \log _{3} 5= \log _{3}(2 x+3⍭ Generate me a similar question◊

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