Solve the equation.
[tex]\begin{array}{c}
\log _6(x-1) + \log _6(x+3)=2\\
x=[?]\\
\end{array}[/tex]
Answer (2)
Combine the logarithms using the property lo g a ( b ) + lo g a ( c ) = lo g a ( b c ) .
Convert the logarithmic equation to an exponential equation.
Solve the resulting quadratic equation using the quadratic formula.
Check the solutions to ensure they are valid within the domain of the logarithmic functions. The final answer is − 1 + 2 10 .
Explanation
Combining Logarithms We are given the equation lo g 6 ( x − 1 ) + lo g 6 ( x + 3 ) = 2 . We need to solve for x . First, we use the logarithm property lo g a ( b ) + lo g a ( c ) = lo g a ( b c ) to combine the logarithms on the left side of the equation.
Converting to Exponential Form Applying the logarithm property, we get lo g 6 (( x − 1 ) ( x + 3 )) = 2 . Next, we convert the logarithmic equation to an exponential equation: ( x − 1 ) ( x + 3 ) = 6 2 .
Expanding and Simplifying Expanding the left side of the equation, we have x 2 + 3 x − x − 3 = 36 . Simplifying the equation, we get x 2 + 2 x − 3 = 36 .
Forming Quadratic Equation Moving all terms to one side to obtain a quadratic equation, we have x 2 + 2 x − 39 = 0 . Now, we solve the quadratic equation for x using the quadratic formula: x = 2 a − b ± b 2 − 4 a c , where a = 1 , b = 2 , and c = − 39 .
Applying Quadratic Formula Plugging in the values, we get x = 2 ( 1 ) − 2 ± 2 2 − 4 ( 1 ) ( − 39 ) = 2 − 2 ± 4 + 156 = 2 − 2 ± 160 = 2 − 2 ± 4 10 = − 1 ± 2 10 . Thus, the two possible solutions are x = − 1 + 2 10 and x = − 1 − 2 10 .
Checking for Validity We need to check the solutions to make sure that 0"> x − 1 > 0 and 0"> x + 3 > 0 , since the logarithm of a non-positive number is undefined. For x = − 1 + 2 10 ≈ − 1 + 2 ( 3.162 ) = − 1 + 6.324 = 5.324 , we have 0"> x − 1 = 4.324 > 0 and 0"> x + 3 = 8.324 > 0 . So, x = − 1 + 2 10 is a valid solution. For x = − 1 − 2 10 ≈ − 1 − 2 ( 3.162 ) = − 1 − 6.324 = − 7.324 , we have x − 1 = − 8.324 < 0 , so x = − 1 − 2 10 is not a valid solution.
Final Solution Therefore, the only solution is x = − 1 + 2 10 .
Examples
Logarithmic equations are used in various fields such as physics, engineering, and finance. For example, in finance, logarithmic scales are used to represent stock market indices or other financial data. Solving logarithmic equations helps in understanding and predicting trends in these areas. Imagine you are analyzing the growth of a bacteria colony. The population doubles every hour, and you want to know how long it will take to reach a certain size. By setting up a logarithmic equation, you can solve for the time it takes for the colony to reach the desired population size. This is a practical application of logarithmic equations in real-world scenarios.
To solve the equation lo g 6 ( x − 1 ) + lo g 6 ( x + 3 ) = 2 , combine the logarithms and convert to exponential form, leading to the quadratic equation x 2 + 2 x − 39 = 0 . After using the quadratic formula, the valid solution is x = − 1 + 2 10 .
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