What is the true solution to $3 \ln 2+\ln 8=2 \ln (4 x) ?$
A. $x=1$
B. $x=2$
C. $x=4$
D. $x=8$
Answer (1)
Simplify the left side of the equation using logarithm properties: 3 ln 2 + ln 8 = ln 64 .
Simplify the right side of the equation using logarithm properties: 2 ln ( 4 x ) = ln ( 16 x 2 ) .
Equate the arguments of the logarithms: 64 = 16 x 2 .
Solve for x : x = 2 .
Explanation
Problem Analysis We are given the equation 3 ln 2 + ln 8 = 2 ln ( 4 x ) and need to find the correct value of x from the given options. We will use properties of logarithms to simplify the equation and solve for x .
Simplifying the Left Side First, we simplify the left side of the equation using the properties of logarithms: 3 ln 2 + ln 8 = ln ( 2 3 ) + ln 8 = ln 8 + ln 8 = ln ( 8 × 8 ) = ln 64 .
Simplifying the Right Side Next, we simplify the right side of the equation using the properties of logarithms: 2 ln ( 4 x ) = ln (( 4 x ) 2 ) = ln ( 16 x 2 ) .
Equating the Arguments Now we have the simplified equation: ln 64 = ln ( 16 x 2 ) .
Since the logarithms are equal, their arguments must be equal: 64 = 16 x 2 .
Solving for x^2 Now, we solve for x 2 :
x 2 = 16 64 = 4 .
Solving for x Taking the square root of both sides, we get: x = ± 2 .
However, since the argument of the logarithm in the original equation is 4 x , we must have 0"> 4 x > 0 , which means 0"> x > 0 . Therefore, we take the positive root: x = 2 .
Final Answer Thus, the true solution to the equation is x = 2 .
Examples
Logarithmic equations are used in various fields such as finance, physics, and engineering. For example, in finance, they are used to calculate the time it takes for an investment to double at a certain interest rate. In physics, they appear in equations describing radioactive decay. Understanding how to solve logarithmic equations is crucial for making informed decisions and solving real-world problems in these areas. Suppose you invest $1000 in an account that pays 5% annual interest, compounded continuously. The amount of time it takes for your investment to double can be found using the equation 2000 = 1000 e 0.05 t , where t is the time in years. Solving for t involves using logarithms.
