/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Solve the equation. $$4^{x} \c... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the equation. $$4^{x} \cdot\left(\frac{1}{2}\right)^{3-2 x}=8 \cdot\left(2^{x}\right)^{2}$$

Short Answer

Expert verified
\(x = 3\)

Step by step solution

01

Simplify the left side

First, express the terms on the left side of the equation with the same base. We begin by rewriting the equation \(4^{x}\) as \((2^2)^x = 2^{2x}\), and \(\left(\frac{1}{2}\right)^{3-2x}\) as \(2^{-(3-2x)} = 2^{2x-3}\). Now, combining these terms, the left side becomes \(2^{2x} \cdot 2^{2x-3} = 2^{4x - 3}\).
02

Simplify the right side

Rewrite the right side of the equation using base 2 for all terms. The term \(8\) is equal to \(2^3\) and \(\left(2^x\right)^2\) becomes \(2^{2x}\). Combining, the right side is \(2^3 \cdot 2^{2x} = 2^{3 + 2x}\).
03

Set the exponents equal

Now that both sides of the equation are expressed as powers of 2, set the exponents equal to each other because if \(2^a = 2^b\), then \(a = b\). Thus, we equate \(4x - 3 = 3 + 2x\).
04

Solve for x

Solve the equation \(4x - 3 = 3 + 2x\). Subtract \(2x\) from both sides: \(2x - 3 = 3\). Then, add 3 to both sides to isolate the term with x: \(2x = 6\). Finally, divide by 2 to find \(x\): \(x = 3\).

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Key Concepts

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

Exponential Functions
Exponential functions are a fascinating part of algebra and mathematics as a whole. They involve equations where the variable appears as an exponent. For instance, in the equation \(4^x\), the base is 4 and the exponent is \(x\). As the value of \(x\) changes, the value of the whole expression changes exponentially, growing or shrinking much faster than simple linear functions.

In real-world applications, exponential functions model a variety of phenomena: from population growth, where the number of individuals multiplies over time, to radioactive decay, where elements decrease by half over specified periods of time. Understanding exponential functions is crucial for grasping more complex scientific and statistical concepts.

Whenever we deal with exponential equations, like the one in the exercise, expressing each part of the equation in terms of the same base helps simplify the problem significantly. This is because comparing exponents is much simpler when all parts are exponential functions with the same base.
Exponent Properties
In order to tackle exponential equations effectively, understanding the properties of exponents is essential. These properties allow us to manipulate and simplify expressions involving powers, making it easier to solve equations.

Let's look at some important rules:
  • Product of Powers Rule: \(a^m \cdot a^n = a^{m+n}\). When multiplying two powers with the same base, you can add the exponents.
  • Power of a Power Rule: \((a^m)^n = a^{mn}\). Here, you multiply the exponents when raising a power to another power.
  • Negative Exponent Rule: \(a^{-n} = \frac{1}{a^n}\). This rule turns a negative exponent into a fraction.
  • Quotient of Powers Rule: \(\frac{a^m}{a^n} = a^{m-n}\). When dividing like bases, subtract the exponents.
In our original exercise, we apply these rules to express both sides of the equation in terms of \(2\) before equating their exponents. This simplifies complex equations and makes them manageable.
Equation Solving
When solving exponential equations, the strategy often involves several calculated steps to express terms with a common base and eliminate challenging exponents. Let's walk through this approach as used in the original scenario.

The first critical step is simplifying the equation either by factoring or using exponent properties to express all terms with the same base, as we did by converting terms to base \(2\). Once you've done that, the equation is easier to handle.

After simplification, set the exponents of the same bases equal, like how if \(2^a = 2^b\), then \(a = b\). This relies on the property of equal bases to directly compare and solve for the variable \(x\).

The final step involves algebraically solving the resulting linear equation by rearranging terms, isolating the variable, and solving for \(x\) in the equation \(4x - 3 = 3 + 2x\), resulting in \(x = 3\).

This method of comparing and adjusting is widely applicable and becomes intuitive with practice, unlocking the power to solve a spectrum of exponential equations.

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Most popular questions from this chapter

Some lending institutions calculate the monthly payment \(M\) on a loan of \(L\) dollars at an interest rate \(r\) (expressed as a decimal) by using the formula $$M=\frac{L r k}{12(k-1)}$$ where \(k=[1+(r / 12)]^{12 t}\) and \(t\) is the number of years that the loan is in effect. Business loan The owner of a small business decides to finance a new computer by borrowing 3000 dollars for 2 years at an interest rate of \(7.5 \%\) (a) Find the monthly payment. (b) Find the total interest paid on the loan.

Exer. \(43-46:\) Use natural logarithms to solve for \(x\) in terms of \(y\) $$y=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$$

One hundred elk, each 1 year old, are introduced into a game preserve. The number \(\bar{N}(t)\) alive after \(t\) years is predicted to be \(N(t)=100(0.9)^{t}\). (a) Estimate the number alive after 5 years. (b) What percentage of the herd dies each year?

Bird calls decrease in intensity (loudness) as they travel through the atmosphere. The farther a bird is from an observer, the softer the sound. This decrease in intensity can be used to estimate the distance between an observer and a bird. A formula that can be used to measure this distance is $$I=I_{0}-20 \log d-k d \text { provided } 0 \leq I \leq I_{0}$$ where \(I_{0}\) represents the intensity (in decibels) of the bird at a distance of one meter (\(I_{0}\) is often known and usually depends only on the type of bird), I is the observed intensity at a distance \(d\) meters from the bird, and \(k\) is a positive constant that depends on the atmospheric conditions such as temperature and humidity. Given \(I_{0}, I,\) and \(k,\) graphically estimate the distance \(d\) between the bird and the observer. $$I_{0}=60, \quad I=15, \quad k=0.11$$

Graph \(f\) on the given interval. (a) Estimate the largest interval \([a, b]\) with \(a< 0< b\) on which \(f\) is one-to-one. (b) If \(g\) is the function with domain \([a, b]\) such that \(g(x)=f(x)\) for \(a \leq x \leq b,\) estimate the domain and range of \(g^{-1}.\) $$f(x)=2.1 x^{3}-2.98 x^{2}-2.11 x+3 ; \quad[-1,2]$$

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