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Chapter 13: Problem 42

Solve each exponential equation. $$32^{y+1}=64^{y+2}$$

Short Answer

Expert verified
The short answer to the given exponential equation \(32^{y+1}=64^{y+2}\) is \(y = -7\).

Step by step solution

01

Rewrite bases as powers of 2

We know that \(32 = 2^5\) and \(64 = 2^6\), so we rewrite the equation using these values: $$\left(2^5\right)^{y+1}=\left(2^6\right)^{y+2}$$
02

Use exponent rules

Now, we apply the exponent rule \((a^m)^n = a^{mn}\) to simplify the equation: $$2^{5(y+1)}=2^{6(y+2)}$$
03

Set the exponents equal to each other

Since both sides of the equation have the same base (2), we can now set the exponents equal to each other: $$5(y+1) = 6(y+2)$$
04

Solve for \(y\)

Now, we will solve for \(y\) using algebraic manipulations: \begin{align*} 5(y+1) &= 6(y+2) \\ 5y+5 &= 6y+12 \\ 5-12 &=6y-5y \\ y &=-7 \end{align*} So, the solution for the given exponential equation is \(y = -7\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponent Rules
When dealing with exponential equations, understanding exponent rules is crucial. These rules help us manipulate and simplify expressions with exponents. Here are some key points about exponent rules you should know:
  • Product of Powers Rule: When multiplying like bases, simply add their exponents. For example, \(a^m \times a^n = a^{m+n}\).
  • Power of a Power Rule: Apply this when you have an exponent raised to another exponent: \((a^m)^n = a^{mn}\). This particular rule is used in solving our exercise.
  • Quotient of Powers Rule: Divide like bases by subtracting the exponents: \(\frac{a^m}{a^n} = a^{m-n}\).
  • Zero Exponent Rule: Any base with an exponent of zero equals one: \(a^0 = 1\), if \(a eq 0\).
Understanding these rules allows you to rewrite and simplify expressions, making it easier to solve exponential equations.
Solving Equations
Solving exponential equations involves finding the value of the variable that makes the equation true. In our exercise, we have the equation \(32^{y+1}=64^{y+2}\). Here鈥檚 how we approach solving it:
  • Rewrite Bases: Start by expressing both sides with a common base. Notice that \(32\) and \(64\) can be rewritten as powers of 2: \(32 = 2^5\) and \(64 = 2^6\).
  • Simplify Using Exponent Rules: Apply the power of a power rule to simplify the equation as so: \((2^5)^{y+1}=(2^6)^{y+2}\) becomes \(2^{5(y+1)}=2^{6(y+2)}\).
  • Set Exponents Equal: Since the bases are now the same (2), set their exponents equal to one another: \(5(y+1) = 6(y+2)\).
These steps help us move forward to find the solution by focusing on the exponents, which is usually more straightforward than solving the entire equation outright.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying equations to solve for an unknown variable. Let's see how we solve for \(y\) in the equation \(5(y+1) = 6(y+2)\):
  • Distribute: Start by using the distributive property: multiply each term inside the parentheses by their respective coefficients to get \(5y + 5 = 6y + 12\).
  • Isolate the Variable: We aim to get all terms containing \(y\) on one side and constants on the other. Subtract \(5y\) from both sides, resulting in \(5 = y + 12\).
  • Simplify Equation: Finally, subtract 12 from both sides to solve for \(y\): \(y = -7\).
By following these steps, we've successfully isolated \(y\), giving us the solution. Understanding these algebraic techniques is essential for tackling various mathematical problems efficiently.

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Most popular questions from this chapter

Find the inverse of each one-to-one function. Then, graph the function and its inverse on the same axes. $$h(x)=-\frac{1}{3} x$$

Find the inverse of each one-to-one function. Then, graph the function and its inverse on the same axes. $$f(x)=-2 x+5$$

Given that \(\log 5=0.6990\) and \(\log 9=0.9542,\) use the propertics of logarithms to approximate the following $$\log \frac{25}{9}$$

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