Question

Multiple Choice Which of the following is a solution of the equation \(2-3^{-x}=-1 ? \mathrm{}\) (A) \(x=-2 \quad\) (B) \(x=-1 \quad\) (C) \(x=0\) (D) \(x=1 \quad\) (E) There are no solutions.

Short answer

The solution to the equation \(2-3^{-x}=-1\) is (C) \(x=0\)

Step-by-step solution

Simplify the equation

Begin by rewriting the equation in a way it will be easier to solve. Add \(3^{-x}\) to both sides of the equation, the equation now becomes \(2 = 3^{-x} + 1\)

Further simplify the equation

Subtract 1 from both sides of the equation, which simplifies the equation to \(1 = 3^{-x}\)

Solve for x

We want to isolate x. To do this, rewrite the equation in logarithmic form using the definition of a logarithm: \(log_{b} a = n \Longrightarrow b^{n} = a\). Rewriting the equation using this definition gives us \(log_{3} 1 = x\). Since any number raised to 0 equals 1, \(x = 0\)

Key concepts

Logarithmic Form

Understanding the logarithmic form is crucial when dealing with equations involving exponents. A logarithm tells us what exponent we need to raise a number to get another number. It's written as \(\log_b(a)\), which answers the question: to what power must we raise \(b\) to obtain \(a\)? When an equation is set up like \(b^x = a\), we can convert it to its logarithmic counterpart: \(\log_{b}(a) = x\). This transformation is essential for solving exponential equations.

For example, if you wanted to solve \(2^x = 8\), converting it to its logarithmic form would give \(\log_{2}(8) = x\), which simplifies to \(x = 3\) because \(2^3 = 8\). In the case of our exercise, the condition \(1 = 3^{-x}\) translates into \(\log_{3}(1) = -x\), making it evident that \(-x = 0\) because \(3^0 = 1\), thereby leading us to the solution that \(x = 0\).

Isolating Variables in Equations

Isolating variables in equations is a fundamental technique in algebra. The goal is to manipulate the equation in a way that you get the unknown variable on one side, making it the subject of the formula. You typically do this through a series of equal operations on both sides of the equation.

For instance, in our exercise, we start with \(2-3^{-x}=-1\). By adding \(3^{-x}\) to both sides, we strategically move all the terms containing \(x\) to one side, resulting in the simpler form \(2=3^{-x}+1\). From here, subtracting 1 from both sides isolates the exponential term, yielding \(1=3^{-x}\). This step-by-step method demonstrates how we incrementally isolate \(x\), preparing it for extraction using the logarithmic form.

Properties of Exponents

The properties of exponents are the toolbox you reach for when handling equations with exponential terms. There are a few key rules, such as the product rule (\(a^m \cdot a^n = a^{m+n}\)), the quotient rule (\(a^m / a^n = a^{m-n}\)), and the power rule (\((a^m)^n = a^{mn}\)), that can simplify complex expressions and aid in solving exponential equations.

In the context of our problem, understanding that \(a^0 = 1\) regardless of what \(a\) is (except when \(a=0\)) leads to the quick realization that the only exponent of 3 that equals 1 would be 0. Consequently, \(-x\) must be 0, and hence \(x = 0\), which matches our choice (C) from the multiple-choice options.

Exponential and Logarithmic Relationships

The relationship between exponential and logarithmic functions is akin to two sides of the same coin. They are inverse functions of each other. This means that logarithms can be used to 'undo' the effect of an exponent, which is a powerful tool when solving exponential equations.

For every equation of the form \(a^x = b\), there is a logarithmic equivalent \(x = \log_a(b)\), which helps us find the value of \(x\). Applying this to our problem, we move from the exponential equation \(1 = 3^{-x}\) to its logarithmic form \(-x = \log_3(1)\), which reveals that \(x = 0\), because the logarithm of 1 to any base is always 0. The relationship between exponents and logarithms thus transforms the complex exponential equation into a simpler logarithmic one, from which we can directly determine the value of the variable \(x\).