Question

Solve for \(x .\) Hint: \(\log _{a} b=c \Leftrightarrow a^{c}=b\). $$ 2 \log _{9}\left(\frac{x}{3}\right)=1 $$

Short answer

The solutions for \(x\) are \(9\) and \(-9\).

Step-by-step solution

Understanding the equation

The given equation is \(2 \log_{9}\left(\frac{x}{3}\right) = 1\). We need to solve for \(x\). To do this, understand that the equation involves a logarithm which we can manage using logarithmic properties.

Applying the logarithm property

The property of logarithms that states \(a \cdot \log_b(c) = \log_b(c^a)\), allows us to rewrite the equation. Using this, the equation becomes \(\log_{9}\left(\left(\frac{x}{3}\right)^2\right) = 1\).

Using the logarithm definition

By the definition of logarithms, \(\log_{9}(d) = c\) means \(9^c = d\). Applying this definition to our equation \(\log_{9}\left(\left(\frac{x}{3}\right)^2\right) = 1\), gives us: \[9^1 = \left(\frac{x}{3}\right)^2\].

Solving the equation

Now solve the equation \(9 = \left(\frac{x}{3}\right)^2\). Begin by taking the square root of both sides, giving \(\frac{x}{3} = \pm \sqrt{9}\). Thus, \(\frac{x}{3} = \pm 3\).

Finding the value of x

To find \(x\), multiply each part of the equation by 3. This yields two possible values for \(x\): \(x = 3 \times 3\) or \(x = 3 \times (-3)\). Hence, \(x = 9\) or \(x = -9\).

Key concepts

Properties of Logarithms

Logarithms have various properties that make them extremely useful for simplifying and solving equations. In the equation given, we focus on the property that states \(a \cdot \log_b(c) = \log_b(c^a)\). This property helps in rewriting expressions where a logarithm is multiplied by a constant. In simpler terms, you can "pull out" a constant multiplier and turn it into an exponent for the argument within the logarithm. In our exercise, this meant transforming \(2 \log_9\left(\frac{x}{3}\right)\) into \(\log_9\left(\left(\frac{x}{3}\right)^2\right)\). Understanding this not only simplifies the problem but also makes it easier to apply further logarithmic concepts, such as solving the equation by using definitions or switching between exponential and logarithmic forms.

Solving Equations

Solving logarithmic equations involves using properties of logarithms to rearrange and simplify the equation into a more manageable form. Once simplified, you can use definitions and algebra to solve for the unknown variable.In our specific task, after applying the logarithmic property, we arrived at the equation \(\log_9\left(\left(\frac{x}{3}\right)^2\right) = 1\). At this point, using the definition of a logarithm, we equate the expression to an exponential form, such as \(9^1 = \left(\frac{x}{3}\right)^2\).This transformation is crucial because exponential equations are often easier to solve than their logarithmic equivalents. You can then use further simplifications, such as taking square roots, to isolate and solve for the unknown.

Logarithms

A logarithm answers the question of how many times a number, called the base, must be multiplied by itself to reach another number. Essentially, if \(\log_b(a) = c\), then \(b^c = a\). This is the fundamental principle underlying logarithms and plays a vital role in solving logarithmic equations.When addressing problems with logarithms, such as the one in our exercise, you often utilize properties to transform them into a form where you can easily apply the definition. In the equation \(2 \log_9\left(\frac{x}{3}\right) = 1\), conversion to exponential form was key to transforming the logarithmic expression into \(9^1 = \left(\frac{x}{3}\right)^2\), making it solvable through algebraic methods.

Square Roots

Square roots are used in equations where a variable is raised to the power of two. The square root function acts as the inverse process, allowing us to find the original number that was squared.In solving the transformed equation \(9 = \left(\frac{x}{3}\right)^2\), taking the square root of both sides is crucial. Doing this simplifies the equation to \(\frac{x}{3} = \pm \sqrt{9}\), which further reduces to \(\frac{x}{3} = \pm 3\).This approach yields two potential solutions: one positive and one negative. Hence, multiplying back the result by 3 provides the two possible values for \(x\), namely 9 and -9. Square roots, therefore, play a key role in breaking down power equations into simpler linear forms.