What Is The Solution To The Equation $4\left(\frac{1}{2}\right)^{x-1}=5x+2$? Round To The Nearest Tenth.A. 0.6 B. 0.7 C. 1.6 D. 5.2
Introduction
Solving equations with exponents can be challenging, but with the right approach, we can find the solution to the given equation. In this article, we will explore the steps to solve the equation $4\left(\frac{1}{2}\right)^{x-1}=5x+2$ and find the value of x.
Understanding the Equation
The given equation is a linear equation with an exponential term. The equation can be rewritten as $2^2 \cdot 2^{-(x-1)}=5x+2$. Using the properties of exponents, we can simplify the equation to $2^{2-(x-1)}=5x+2$.
Simplifying the Equation
To simplify the equation, we can rewrite the exponent as $2^{3-x}=5x+2$. This equation can be further simplified by using the fact that $2{3-x}=\frac{23}{2x}=\frac{8}{2x}$.
Isolating the Exponential Term
Now, we can rewrite the equation as $\frac{8}{2^x}=5x+2$. To isolate the exponential term, we can multiply both sides of the equation by $2^x$, resulting in $8=(5x+2)2^x$.
Using Logarithms to Solve the Equation
To solve the equation, we can use logarithms to eliminate the exponential term. Taking the logarithm of both sides of the equation, we get $\log(8)=(x+2)\log(2)+\log(5x+2)$.
Simplifying the Logarithmic Equation
Using the properties of logarithms, we can simplify the equation to $\log(8)=x\log(2)+2\log(2)+\log(5x+2)$. This equation can be further simplified by combining like terms, resulting in $\log(8)=x\log(2)+\log(5x+2)+2\log(2)$.
Isolating the Variable
To isolate the variable x, we can subtract $2\log(2)$ from both sides of the equation, resulting in $\log(8)-2\log(2)=x\log(2)+\log(5x+2)$. This equation can be further simplified by combining like terms, resulting in $\log(4)=x\log(2)+\log(5x+2)$.
Using Properties of Logarithms
Using the properties of logarithms, we can rewrite the equation as $\log(4)=\log(2^x)+\log(5x+2)$. This equation can be further simplified by combining the logarithmic terms, resulting in $\log(4)=\log(2^x(5x+2))$.
Equating the Arguments of the Logarithms
Since the logarithms are equal, we can equate the arguments of the logarithms, resulting in $4=2^x(5x+2)$. This equation can be further simplified by dividing both sides of the equation by 4, resulting in $1=\frac{2^x(5x+2)}{4}$.
Simplifying the Equation
To simplify the equation, we can multiply both sides of the equation by 4, resulting in $4=2^x(5x+2)$. This equation can be further simplified by dividing both sides of the equation by 2, resulting in $2=2^{x-1}(5x+2)$.
Using the Definition of Exponentiation
Using the definition of exponentiation, we can rewrite the equation as $2=2^{x-1}(5x+2)$. This equation can be further simplified by dividing both sides of the equation by 2, resulting in $1=2^{x-2}(5x+2)$.
Solving for x
To solve for x, we can divide both sides of the equation by $2^{x-2}$, resulting in $\frac{1}{2^{x-2}}=5x+2$. This equation can be further simplified by multiplying both sides of the equation by $2^{x-2}$, resulting in $1=(5x+2)2^{x-2}$.
Using Logarithms to Solve for x
To solve for x, we can use logarithms to eliminate the exponential term. Taking the logarithm of both sides of the equation, we get $\log(1)=(x+2)\log(2)+\log(5x+2)$.
Simplifying the Logarithmic Equation
Using the properties of logarithms, we can simplify the equation to $0=x\log(2)+2\log(2)+\log(5x+2)$. This equation can be further simplified by combining like terms, resulting in $0=x\log(2)+\log(5x+2)+2\log(2)$.
Isolating the Variable
To isolate the variable x, we can subtract $2\log(2)$ from both sides of the equation, resulting in $-2\log(2)=x\log(2)+\log(5x+2)$. This equation can be further simplified by combining like terms, resulting in $-2\log(2)=x\log(2)+\log(5x+2)$.
Using Properties of Logarithms
Using the properties of logarithms, we can rewrite the equation as $-2\log(2)=\log(2^x)+\log(5x+2)$. This equation can be further simplified by combining the logarithmic terms, resulting in $-2\log(2)=\log(2^x(5x+2))$.
Equating the Arguments of the Logarithms
Since the logarithms are equal, we can equate the arguments of the logarithms, resulting in $2{-2}=2x(5x+2)$. This equation can be further simplified by dividing both sides of the equation by 2, resulting in $2{-1}=2{x-1}(5x+2)$.
Solving for x
To solve for x, we can divide both sides of the equation by $2^{x-1}$, resulting in $\frac{1}{2}=5x+2$. This equation can be further simplified by subtracting 2 from both sides of the equation, resulting in $-1=5x$.
Finding the Value of x
To find the value of x, we can divide both sides of the equation by 5, resulting in $x=-\frac{1}{5}$.
Rounding to the Nearest Tenth
To round the value of x to the nearest tenth, we get $x=-0.2$.
Conclusion
In this article, we have explored the steps to solve the equation $4\left(\frac{1}{2}\right)^{x-1}=5x+2$. We have used logarithms to eliminate the exponential term and have found the value of x to be $x=-0.2$. This value is rounded to the nearest tenth.
Final Answer
The final answer is:
Introduction
In our previous article, we explored the steps to solve the equation $4\left(\frac{1}{2}\right)^{x-1}=5x+2$. We used logarithms to eliminate the exponential term and found the value of x to be $x=-0.2$. In this article, we will answer some of the most frequently asked questions about solving this equation.
Q: What is the first step in solving the equation $4\left(\frac{1}{2}\right)^{x-1}=5x+2$?
A: The first step in solving the equation is to rewrite the equation as $2^2 \cdot 2^{-(x-1)}=5x+2$. This simplifies the equation and makes it easier to work with.
Q: How do you simplify the equation $2^2 \cdot 2^{-(x-1)}=5x+2$?
A: To simplify the equation, we can use the properties of exponents to rewrite the equation as $2^{2-(x-1)}=5x+2$. This simplifies the equation and makes it easier to work with.
Q: What is the next step in solving the equation $2^{2-(x-1)}=5x+2$?
A: The next step in solving the equation is to use logarithms to eliminate the exponential term. We can take the logarithm of both sides of the equation to get $\log(2^{2-(x-1)})=\log(5x+2)$.
Q: How do you use logarithms to eliminate the exponential term in the equation $\log(2^{2-(x-1)})=\log(5x+2)$?
A: To use logarithms to eliminate the exponential term, we can use the property of logarithms that states $\log(a^b)=b\log(a)$. We can apply this property to the equation to get $(2-(x-1))\log(2)=\log(5x+2)$.
Q: What is the next step in solving the equation $(2-(x-1))\log(2)=\log(5x+2)$?
A: The next step in solving the equation is to isolate the variable x. We can do this by subtracting $2\log(2)$ from both sides of the equation to get $\log(4)-2\log(2)=x\log(2)+\log(5x+2)$.
Q: How do you isolate the variable x in the equation $\log(4)-2\log(2)=x\log(2)+\log(5x+2)$?
A: To isolate the variable x, we can subtract $\log(5x+2)$ from both sides of the equation to get $\log(4)-2\log(2)-\log(5x+2)=x\log(2)$. We can then divide both sides of the equation by $\log(2)$ to get $\frac{\log(4)-2\log(2)-\log(5x+2)}{\log(2)}=x$.
Q: What is the final step in solving the equation $\frac{\log(4)-2\log(2)-\log(5x+2)}{\log(2)}=x$?
A: The final step in solving the equation is to evaluate the expression on the left-hand side of the equation. We can do this by using a calculator to find the value of the expression. This will give us the value of x.
Q: What is the value of x in the equation $\frac{\log(4)-2\log(2)-\log(5x+2)}{\log(2)}=x$?
A: The value of x in the equation is $x=-0.2$.
Q: How do you round the value of x to the nearest tenth?
A: To round the value of x to the nearest tenth, we can look at the hundredth place value of the number. If the hundredth place value is 5 or greater, we round up. If the hundredth place value is less than 5, we round down. In this case, the hundredth place value of the number is 2, so we round down to get $x=-0.2$.
Conclusion
In this article, we have answered some of the most frequently asked questions about solving the equation $4\left(\frac{1}{2}\right)^{x-1}=5x+2$. We have used logarithms to eliminate the exponential term and have found the value of x to be $x=-0.2$. This value is rounded to the nearest tenth.
Final Answer
The final answer is: