Problem 70
Question
Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(5 \log _{10}(x-2)=11\)
Step-by-Step Solution
Verified Answer
The exact solution is \(x=10^{\frac{11}{5}}+2\). To find an approximate solution, use a calculator to compute the value of \(10^{\frac{11}{5}}\) and then add 2. This gives approximately \(x=50.932\).
1Step 1: Isolate the logarithm
First, divide the entire equation by 5 to isolate the logarithm. This gives us: \(\log _{10}(x-2)=\frac{11}{5}\).
2Step 2: Convert to exponential form
The equation can then be rewritten into exponential form to further isolate \(x\). This is done based on the definition of a logarithm: if \(a=log_b(c)\), then \(b^a=c\). Applying this to our equation gives: \(10^{\frac{11}{5}}=x-2\).
3Step 3: Solve for \(x\)
Lastly, solve for \(x\) by adding 2 to both sides of the equation: \(x=10^{\frac{11}{5}}+2\).
Key Concepts
Isolate the LogarithmConvert Logarithm to Exponential FormAlgebraic Manipulation
Isolate the Logarithm
Understanding how to isolate a logarithm is fundamental when solving logarithmic equations. In the given problem, we start with the equation
When isolating the logarithm, remember that your goal is to have the logarithm on one side of the equation with a coefficient of 1. This simplifies the equation and prepares it for conversion to exponential form. Being patient and careful with your algebra here can prevent mistakes that might complicate the problem as you progress.
5 \times \text{log}_{10}(x - 2) = 11. To isolate the logarithm, we divide both sides of the equation by 5, which yields \( \text{log}_{10}(x - 2) = \frac{11}{5} \). This step is crucial because it allows the logarithm to stand alone, making the next steps more straightforward.When isolating the logarithm, remember that your goal is to have the logarithm on one side of the equation with a coefficient of 1. This simplifies the equation and prepares it for conversion to exponential form. Being patient and careful with your algebra here can prevent mistakes that might complicate the problem as you progress.
Convert Logarithm to Exponential Form
Once the logarithm is isolated, as in our simplified equation \( \text{log}_{10}(x - 2) = \frac{11}{5} \), the next step is to convert it into exponential form. Logarithms and exponentials are inverse functions, which means that converting a log to its exponential counterpart allows us to solve for the variable. According to the definition, if \( a = \text{log}_{b}(c) \), then the exponential form is \( b^{a} = c \).
For our problem, converting \( \text{log}_{10}(x - 2) = \frac{11}{5} \) to exponential form gives us \( 10^{\frac{11}{5}} = x - 2 \). Always ensure that the base in the exponential form matches the base of the logarithm—in this case, it's 10, which is a common base for logarithms. Converting to exponential form helps us to see the problem from a different angle and gets us one step closer to finding the value of x.
For our problem, converting \( \text{log}_{10}(x - 2) = \frac{11}{5} \) to exponential form gives us \( 10^{\frac{11}{5}} = x - 2 \). Always ensure that the base in the exponential form matches the base of the logarithm—in this case, it's 10, which is a common base for logarithms. Converting to exponential form helps us to see the problem from a different angle and gets us one step closer to finding the value of x.
Algebraic Manipulation
Algebraic manipulation is essential for solving equations, including logarithmic equations. After converting to exponential form, we have the equation \( 10^{\frac{11}{5}} = x - 2 \). The next task is to solve for \( x \), which requires further algebraic manipulation. We add 2 to both sides of the equation to get \( x = 10^{\frac{11}{5}} + 2 \).
This final step exemplifies algebraic manipulation, which involves moving terms from one side of an equation to the other to isolate the variable we're solving for. It's like a balancing act—what you do to one side, you must do to the other to maintain equality. Algebraic manipulation will vary based on the problem but might involve techniques like adding, subtracting, multiplying, dividing, or even factoring. In our case, a simple addition was all that was needed to find the solution to the logarithmic equation.
This final step exemplifies algebraic manipulation, which involves moving terms from one side of an equation to the other to isolate the variable we're solving for. It's like a balancing act—what you do to one side, you must do to the other to maintain equality. Algebraic manipulation will vary based on the problem but might involve techniques like adding, subtracting, multiplying, dividing, or even factoring. In our case, a simple addition was all that was needed to find the solution to the logarithmic equation.
Other exercises in this chapter
Problem 69
Find the domain, vertical asymptote, and \(x\) -intercept of the logarithmic function. Then sketch its graph.\(f(x)=-\log _{2} x\)
View solution Problem 69
Radioactive Decay Five pounds of the element plutonium \(\left({ }^{230} \mathrm{Pu}\right)\) is released in a nuclear accident. The amount of plutonium \(P\) t
View solution Problem 70
Use the properties of logarithms to expand the expression as a sum, difference, and/or multiple of logarithms. (Assume all variables are positive.)\(\log _{10}
View solution Problem 70
Find the domain, vertical asymptote, and \(x\) -intercept of the logarithmic function. Then sketch its graph.\(h(x)=-\log _{4}(x-1)\)
View solution