About Solving Logarithmic Equations:
When working with logarithmic equations, we utilize a variety of techniques to obtain a solution. In some cases, we will need to convert into exponential form to get a solution. In other cases, we will rely on some basic properties of logarithms to obtain our solution.
Test Objectives
- Demonstrate the ability to convert between exponential and logarithmic form
- Demonstrate the ability to Solve Logarithmic Equations
#1:
Instructions: solve each equation.
$$a)\hspace{.2em}-6log_{9}(-9x-6)-9=-15$$
$$b)\hspace{.2em}2log_{3}(-6x-5)-9=-13$$
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#2:
Instructions: solve each equation.
$$a)\hspace{.2em}9log_{7}(-6x-5)-10=-19$$
$$b)\hspace{.2em}log_{6}(-13x-3)=log_{6}(x^2 + 27)$$
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#3:
Instructions: solve each equation.
$$a)\hspace{.2em}log_{7}(8x + 1)=log_{7}(x^2 + 1)$$
$$b)\hspace{.2em}log(3 - 4x) - log(7)=1$$
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#4:
Instructions: solve each equation.
$$a)\hspace{.2em}log_{3}(3 - 5x) + log_{3}(7)=log_{3}(24)$$
$$b)\hspace{.2em}log_{5}(4x^2 + 7) - log_{5}(9)=1$$
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#5:
Instructions: solve each equation.
$$a)\hspace{.2em}ln(x - 13) + ln(x - 3)=ln(75)$$
$$b)\hspace{.2em}ln(x + 3) - ln(x - 1)=ln(4)$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}x=-\frac{5}{3}$$
$$b)\hspace{.2em}x=-\frac{23}{27}$$
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#2:
Solutions:
$$a)\hspace{.2em}x=-\frac{6}{7}$$
$$b)\hspace{.2em}x=-3, -10$$
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#3:
Solutions:
$$a)\hspace{.2em}x=8,0$$
$$b)\hspace{.2em}x=-\frac{67}{4}$$
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#4:
Solutions:
$$a)\hspace{.2em}x=-\frac{3}{35}$$
$$b)\hspace{.2em}x=\frac{\pm \sqrt{38}}{2}$$
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#5:
Solutions:
$$a)\hspace{.2em}x=18$$
$$b)\hspace{.2em}x=\frac{7}{3}$$
