Problem 17 - Precalculus : Building Concepts and Connections

Question

Solve the exponential equation. Round to three decimal places, when needed. $$10^{x}=2^{-x+4}$$

Step-by-Step Solution

Answer

The solution of the equation is approximately x ≈ 1.514.

Step 1: Apply logarithms to both sides of the equation

To get rid of the exponents, apply the logarithm to both sides of the equation, it implies using the natural logarithm: \[\ln(10^{x}) = \ln(2^{-x+4})\]By the rule of logarithms, the coefficient of the logarithm can come out front as a multiplication, so the equation becomes:\[x\ln(10) = (-x+4) \ln(2)\]

Step 2: Bring all terms with x to one side of the equation

We want to isolate x, so we need to bring all terms that include x to one side. Rearrange above equation to have x factors on one side:\[x\ln(10) + x\ln(2) = 4\ln(2)\]Or, taking x as common:\[x(\ln(10) + \ln(2)) = 4\ln(2)\]

Step 3: Solve for x

Now we can isolate x for our solution. Divide both sides of the equation by \((\ln(10) + \ln(2))\). We get:\[x = \frac{4\ln(2)}{\ln(10) + \ln(2)}\]To find x with the prescribed three decimal places, use a calculator to find x ≈ 1.514

Key Concepts

Understanding Logarithmic Functions

Logarithmic functions play a crucial role in solving exponential equations as they allow us to "bring down" exponents, making the equation more manageable. A logarithm is essentially the inverse of an exponential function, meaning that if you know one, you can determine the other.
For example, given the expression \(10^x\), applying a logarithm converts it to \(\log_{10}(10^x) = x\cdot \log_{10}(10)\). Similarly, for the expression \(2^{-x+4}\), applying a logarithm results in \((-x+4)\cdot \log_{10}(2)\).
Thus, logarithms allow us to transform complex exponential equations into linear equations, which are considerably easier to solve. Different bases of logarithms, such as common logarithm (base 10, written as \(\log\)) or natural logarithm (base \(e\), written as \(\ln\)), can be used based on the problem requirement. Each logarithm type simplifies the calculations according to their respective base.
For this exercise, we turned to the natural logarithm to harness its consistent base \(e\) and straightforward application in scientific computations.

The Role of Natural Logarithm

The natural logarithm, often expressed as \(\ln\), is used extensively in mathematics and sciences because of its unique properties that align elegantly with the nature of exponential growth. Specifically, \(\ln\) uses the base \(e\), which is an irrational constant approximately equal to 2.718.
When applied to an exponential equation, the natural logarithm simplifies the process of resolving variables nestled within exponents. For instance, in the given equation \(10^{x} = 2^{-x+4}\), using \(\ln\):

• First, take the natural logarithm of both sides, giving: \(\ln(10^{x}) = \ln(2^{-x+4})\).
• By logarithmic rules, bring the exponents down as multipliers: \(x \cdot \ln(10) = (-x+4) \cdot \ln(2)\).

This simplifies to linear terms, making the equation solvable via ordinary algebraic manipulation.
The decision to use \(\ln\) can be influenced by requirements of the equation or preferred simplicity. Whether \(\log\) or \(\ln\), the use of logarithms in general aids in isolating and solving for variables that might otherwise be challenging to manage.

Steps to Isolating Variables

Isolating variables is a vital step in solving equations, especially those involving exponential forms and logarithms. Once logarithmic functions have simplified an equation, isolating the variable typically involves straightforward algebraic operations.
In this exercise, after applying logarithms, the equation was rewritten to:

• \(x \cdot \ln(10) + x \cdot \ln(2) = 4 \cdot \ln(2)\)
• Factor out \(x\) to have it on one side: \(x(\ln(10) + \ln(2)) = 4\ln(2)\).

The goal here is to solve for \(x\). By dividing both sides by \(\ln(10) + \ln(2)\), you isolate \(x\):
\[x = \frac{4 \ln(2)}{\ln(10) + \ln(2)}\]
This methodical approach to isolating variables ensures clear and concise solutions. Performing each algebraic step carefully, and sometimes breaking them into smaller parts, helps to avoid mistakes and aligns solutions closely with correct mathematical processes.