Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(3\left(5^{x-1}\right)=21\)

Short answer

The solution to the equation is \(x = 1.210\)

Step-by-step solution

Isolate the Exponential

In the equation, \(3\left(5^{x-1}\right)=21\), divide both sides by 3 to isolate the exponential term. We get \(5^{x-1} = 7\)

Take Natural Logarithm of Both Sides

Applying the natural logarithm to both sides to handle the exponential, we get \(\ln(5^{x-1}) = \ln(7)\)

Simplify

Using a property of logarithms, namely \(\log_b(a^n) = n \log_b(a)\), we can pull down the exponent and multiply the logarithm by it. We get \((x-1)\ln(5) = \ln(7)\)

Solve for x

Isolate \(x\) by dividing both sides by \(\ln(5)\), then add 1. This yields the solution \(x = \frac{\ln(7)}{\ln(5)} + 1 \)

Approximate the Result

Now, use the natural logarithmic values of 7 and 5 to calculate the value of x to three decimal places: \(x = \frac{1.94591014906}{1.60943791243} + 1 = 1.210\).

Key concepts

Algebraic Manipulation

Understanding how to rearrange and simplify equations is fundamental to solving exponential equations. Algebraic manipulation involves using math operations to isolate the unknown variable. In our example, the initial equation is given as \(3(5^{x-1}) = 21\). The goal is to solve for \(x\).

To begin, you need to isolate the exponential term by performing operations that make the equation simpler to handle. This means dividing both sides by 3 to eliminate the coefficient next to the exponential term. The resulting equation is \(5^{x-1} = 7\), which sets the stage for the next steps of introducing logarithms to solve for \(x\).

Properties of Logarithms

Logarithms are invaluable tools when dealing with exponential equations. The properties of logarithms allow you transform an exponential equation into a form that is much easier to solve algebraically. One such property that's particularly useful is that \(\log_b(a^n) = n\log_b(a)\), where \(a\) is the base, \(b\) is the logarithm's base, and \(n\) is the exponent.

In the solution process, after taking the natural logarithm of both sides of the equation \(5^{x-1} = 7\), we apply this property to move the exponent \((x-1)\) in front of the logarithm, which significantly simplifies the expression to \((x-1)\ln(5) = \ln(7)\). This step is crucial for the next phase of solving for \(x\).

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is a specific type of logarithm with the base \(e\), where \(e\) is the Euler's number (approximately 2.71828). It is widely used due to how naturally it arises in mathematical contexts, especially in calculus and complex equations involving growth and decay.

When we deal with exponential equations such as \(5^{x-1} = 7\), taking the natural logarithm of both sides makes the equation manageable since the exponent can now be isolated. By using the natural logarithm and the logarithmic property mentioned earlier, we can solve for \(x\) easily. In our exercise, this transformation leads to a clear path for isolating and solving for the variable \(x\).

Approximating Solutions

After we have an isolated \(x\) in the form of \(x = (\ln(7))/(\ln(5)) + 1\), we need to find its approximate value. In many cases, the exact solutions are complex or not possible to express in simplified form. Therefore, we rely on approximation techniques to find a numerical solution to a degree of precision.

In mathematics, digital calculators or software efficiently find the values of the natural logarithms of numbers to several decimal places. For our example, the values of \(\ln(7)\) and \(\ln(5)\) are approximated and then used to calculate \(x\) to three decimal places, resulting in \(x = 1.210\). Approximation is a key skill as it allows the expression of solutions to real-world problems where exact values are not necessary or are cumbersome to work with.