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Chapter 5: Problem 55

Solve each equation. $$6^{1-x}=\left(\frac{1}{36}\right)^{2 x}$$

Short Answer

Expert verified
The solution is \(x = -\frac{1}{3}\).

Step by step solution

01

Express Both Sides with the Same Base

First, simplify the equation so that both sides have the same base. We have the equation:\[ 6^{1-x} = \left(\frac{1}{36}\right)^{2x} \]We know that \(36 = 6^2\), so \(\frac{1}{36} = (6^2)^{-1} = 6^{-2}\). Substituting gives:\[ \left(\frac{1}{36}\right)^{2x} = (6^{-2})^{2x} = 6^{-4x} \]
02

Equate the Exponents

Now that both sides of the equation have the same base, we can set the exponents equal to each other:\[ 1 - x = -4x \]This is because the equation \(6^{1-x} = 6^{-4x}\) implies that the exponents must be equal when the bases are the same.
03

Solve the Exponent Equation

Solve the equation \(1 - x = -4x\) for \(x\):Add \(x\) to both sides to isolate \(x\) on one side:\[ 1 = -4x + x \]\[ 1 = -3x \]Now divide both sides by \(-3\):\[ x = -\frac{1}{3} \]
04

Verify the Solution

Substitute \(x = -\frac{1}{3}\) back into the original equation to verify the solution:The left side becomes:\[ 6^{1-(-\frac{1}{3})} = 6^{1+\frac{1}{3}} = 6^{\frac{4}{3}} \]The right side becomes:\[ \left(\frac{1}{36}\right)^{2(-\frac{1}{3})} = \left(\frac{1}{36}\right)^{-\frac{2}{3}} = 6^{\frac{4}{3}} \]Both sides are equal, confirming that \(x = -\frac{1}{3}\) is the correct solution.

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Key Concepts

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

Solving Equations
Solving equations is a fundamental aspect of algebra. It involves finding the value of the variable that makes the equation true. In our exercise, we start with the equation \(6^{1-x}=\left(\frac{1}{36}\right)^{2x}\). The goal is to find the value of \(x\) that satisfies this equation.
To simplify our task, we first need to rewrite both sides of the equation with a common base. This step makes it easier to compare the exponents directly. Once we've expressed both sides with the same base, we can ignore the bases and focus solely on the exponents, equating them to one another. This step reduces the equation to a simpler form that is easier to handle.
  • The first step in solving is to ensure both sides of the equation have the same base.
  • After simplifying, compare the exponents directly.
  • Find the value of the variable by solving the simpler equation.
In essence, solving equations involves logical steps such as simplifying expressions and systematically solving for the variable by isolating it.
Exponents
Exponents are a mathematical notation that expresses the number of times a number (base) is multiplied by itself. In our task, the equation \(6^{1-x}=\left(\frac{1}{36}\right)^{2x}\) involves exponents. To solve it, understanding how to manipulate exponents is essential.
In the given equation, \( \left(\frac{1}{36}\right)^{2x} \) can be rewritten using a negative exponent. Recall that any fraction raised to a power can be converted by finding the reciprocal and changing the sign of the exponent. Hence, \(\frac{1}{36} = 36^{-1}\) and equivalently \(6^{-2}\) because \(36 = 6^2\). These manipulations help us rewrite the expression with a common base of \(6\).
  • Exponents determine how many times the base is multiplied by itself.
  • Using negative exponents simplifies reciprocals within expressions.
  • Identical bases allow us to relate exponents directly.
Handling exponents skillfully allows the equation to be simplified, facilitating easier solution of the variable.
Algebraic Manipulation
Algebraic manipulation is the technique of rearranging and adjusting equations to solve for an unknown variable. In dealing with the equation \(6^{1-x}=6^{-4x}\), we've leveraged algebraic manipulation at various stages.
Once both sides have a common base, we equate the exponents, hence simplifying the equation to \(1-x=-4x\). This provides a straightforward linear equation where we isolate the variable \(x\). By adding \(x\) to both sides, we systematically work towards isolating \(x\). Finally, solving for \(x\) involves basic arithmetic operations like division.
  • Equating exponents is a form of algebraic manipulation when the bases are identical.
  • Isolating the variable involves adding, subtracting, and dividing terms strategically.
  • Verification step ensures our manipulated solution satisfies the original equation.
Algebraic manipulation is key to transforming complex equations into simpler forms that are more manageable to solve.

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Most popular questions from this chapter

Emissions Governments could reduce carbon emissions by placing a tax on fossil fuels. The costbenefit equation $$\ln (1-P)=-0.0034-0.0053 x$$ estimates the relationship between a tax of \(x\) dollars per ton of carbon and the percent \(P\) reduction in emissions of carbon, where \(P\) is in decimal form. Determine \(P\) when \(x=60 .\) Interpret the result. (Source: Clime, W., The Economics of Global Warming, Institute for International Economics.)

Sprinter's Speed and Time During the 100 -meter dash, the elapsed time \(T\) for a sprinter to reach a speed of x meters per second is given by the following function. $$T(x)=-1.2 \ln \left(1-\frac{x}{11}\right)$$ (a) How much time had elapsed when the sprinter was running 0 meters per second? Interpret your answer. (b) At the end of the race, the sprinter was moving at 10.998 meters per second. What was the sprinter's time for this 100 -meter dash? (c) Find T^{-1}(x) and interpret its meaning.

Escherichia coli is a strain of bacteria that occurs naturally in many organisms. Under certain conditions, the number of bacteria present in a colony is approximated by $$A(t)=A_{0} e^{0.023 t}$$ where \(t\) is in minutes. If \(A_{0}=2,400,000,\) find the number of bacteria at each time. Round to the nearest hundred thousand. (a) 5 minutes (b) 10 minutes (c) 60 minutes

Newton's law of cooling says that the rate at which an object cools is proportional to the difference \(C\) in temperature between the object and the environment around it. The temperature \(f(t)\) of the object at time t in appropriate units after being introduced into an environment with a constant temperature \(T_{0}\) is $$f(t)=T_{0}+C e^{-k t}$$ where \(C\) and \(k\) are constants. Use this result. Boiling water at \(100^{\circ} \mathrm{C}\) is placed in a freezer at \(0^{\circ} \mathrm{C}\). The temperature of the water is \(50^{\circ} \mathrm{C}\) after 24 minutes. Approximate the temperature of the water after 96 minutes.

Suppose that the concentration of a bacteria sample is \(100,000\) bacteria per milliliter. If the concentration doubles every 2 hours, how long will it take for the concentration to reach \(350,000\) bacteria per milliliter?
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