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Chapter 3: Problem 20

Solve the equation for \(x\). $$3^{x-x^{2}}=\frac{1}{9^{x}}$$

Short Answer

Expert verified
The short answer is: rewrite the given equation as \(3^{x-x^{2}}=3^{-2x}\), then set the exponents equal, leading to the quadratic equation \(x^2-3x=0\). Factor the equation and solve for \(x\), giving the solutions \(x=0\) and \(x=3\).

Step by step solution

01

Rewrite the equation with the same base on both sides.

Since \(9\) is equal to \(3^2\), we can rewrite the equation with the same base of 3 on both sides: \[3^{x-x^{2}}=\frac{1}{3^{2x}}\]
02

Simplify the equation.

Now we can simplify the equation by getting rid of the fraction on the right side: \[3^{x-x^{2}}=3^{-2x}\]
03

Compare the exponents of the equation.

Since the bases are equal, the exponents must be equal as well. So, we have: \[x-x^{2}=-2x\]
04

Solve for x by making the quadratic equation.

If we rearrange the equation, we get a quadratic equation: \[x^2-3x=0\]
05

Factor the equation and solve for x.

The quadratic equation can be factored as: \[x(x-3)=0\] So we have two possible solutions for \(x\): \[x=0\] or \[x=3\] Hence, the solutions for the given equation are \(x=0\) and \(x=3\).

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Key Concepts

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

Quadratic Equations
Quadratic equations are polynomials of degree two, meaning they are represented by the form \(ax^2 + bx + c = 0\). The standard method to solve quadratic equations is by factoring, using the quadratic formula, or completing the square. In our exercise, the equation \(x^2 - 3x = 0\) is derived. This is a simpler form of a quadratic equation since it misses the constant term \(c\). For this specific equation, it is easier to solve by factoring:
  • First, set the quadratic expression equal to zero: \(x^2 - 3x = 0\).
  • Next, factor out the common term \(x\), which gives you \(x(x - 3) = 0\).
  • According to the zero-product property, you set each factor equal to zero: \(x = 0\) or \(x - 3 = 0\).
  • Solve for \(x\) in each case, resulting in \(x = 0\) and \(x = 3\).
The solutions for \(x\) are the points where the graph of the equation crosses the x-axis. These solutions are significant in understanding how quadratic equations model relationships where results can have multiple solutions.
Base Conversion
Base conversion is about expressing a number in a different base or form that is more convenient for mathematical operations. In the context of exponential equations, this usually involves finding a common base to simplify calculations. Let's look at the exercise: it includes expressions with bases that need alignment 鈥� converting numbers to the same base makes it easier to compare or solve them.
  • The number 9 can be re-expressed as \(3^2\) because \(9 = 3 \times 3\).
  • As a crucial step, when solving \(3^{x-x^2} = \frac{1}{9^x}\), you need both sides in the base of 3.
  • Convert \(\frac{1}{9^x}\) to \(3^{-2x}\), understanding that the exponent changes sign when a term is moved from denominator to numerator in a fraction.
This step ensures that the equation can be rewritten and simplified without losing precision, making further steps more intuitive. Base conversion helps harness properties of exponents, like equalizing and simplifying exponents when the bases are unified.
Exponent Comparison
Exponent comparison is key in solving exponential equations once bases are identical. When both sides of an equation are rewritten to have the same base, it allows us to directly compare and equate the exponents.In our step-by-step solution, once the equation is simplified to \(3^{x-x^{2}} = 3^{-2x}\), you can proceed to compare the exponents:
  • If \(a^m = a^n\) for any base \(a\) not equal to zero, it follows that \(m = n\).
  • Therefore, the equation \(x-x^{2} = -2x\) follows directly from setting the exponents equal.
  • Simplifying \(x-x^{2} = -2x\) by adding \(2x\) to both sides yields the quadratic \(x^2 - 3x = 0\).
This methodology leverages the power and convenience of working with exponential functions and simplifies complex comparisons into more basic algebraic manipulations. Exponent comparison is a powerful tool in any equation-solving toolkit, especially when determining identical equations or solving for unknown variables.

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Most popular questions from this chapter

a. Given that \(2^{x}=e^{k x}\), find \(k\). b. Show that, in general, if \(b\) is a nonnegative real number, then any equation of the form \(y=b^{x}\) may be written in the form \(y=e^{k x}\), for some real number \(k\).

Metro Department Store found that \(t\) wk after the end of a sales promotion the volume of sales was given by $$ S(t)=B+A e^{-\lambda t} \quad(0 \leq t \leq 4) $$ where \(B=50,000\) and is equal to the average weekly volume of sales before the promotion. The sales volumes at the end of the first and third weeks were \(\$ 83,515\) and \(\$ 65,055\), respectively. Assume that the sales volume is decreasing exponentially. a. Find the decay constant \(k\). b. Find the sales volume at the end of the fourth week.

Use logarithms to solve the equation for \(t\). $$\frac{A}{1+B e^{t / 2}}=C$$

Halley's law states that the barometric pressure (in inches of mercury) at an altitude of \(x \mathrm{mi}\) above sea level is approximated by the equation $$ p(x)=29.92 e^{-0.2 x} \quad(x \geq 0) $$ If the barometric pressure as measured by a hot-air balloonist is 20 in. of mercury, what is the balloonist's altitude?

The length (in centimeters) of a typical Pacific halibut \(t\) yr old is approximately $$ f(t)=200\left(1-0.956 e^{-0.18 t}\right) $$ What is the length of a typical 5 -yr-old Pacific halibut?
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