Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. See Examples 1–4. 6^(x+1) = 4^(2x-1)

Verified step by step guidance

1. Take the logarithm of both sides of the equation: $\log (6^{x+1})=\log (4^{2x-1})$.
2. Use the power rule of logarithms to bring down the exponents: $(x+1)\log (6)=(2x-1)\log (4)$.
3. Distribute the logarithms: $x\log (6)+\log (6)=2x\log (4)-\log (4)$.
4. Rearrange the equation to isolate terms involving $x$: $x\log (6)-2x\log (4)=-\log (6)-\log (4)$.
5. Factor out $x$ from the left side: $x(\log (6)-2\log (4))=-\log (6)-\log (4)$.

Key Concepts

Exponential Equations

Exponential equations involve variables in the exponent, such as the equation 6^(x+1) = 4^(2x-1). To solve these, one often uses properties of exponents, logarithms, or by rewriting the bases to be the same. Understanding how to manipulate these equations is crucial for finding the value of x.

Logarithms

Logarithms are the inverse operations of exponentiation and are essential for solving exponential equations. For example, if you have an equation like a^b = c, you can take the logarithm of both sides to isolate the variable. This concept allows for the transformation of exponential forms into linear forms, making them easier to solve.

Exact vs. Approximate Solutions

In mathematics, exact solutions are expressed in their simplest form, while approximate solutions are numerical values rounded to a specified degree of accuracy. In this problem, you are asked to provide irrational solutions as decimals to the nearest thousandth and exact forms for other exercises, highlighting the importance of understanding when to use each type of solution.