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Chapter 3: Problem 14

Solve for \(x\) algebraically. $$10^{6-x}=550$$

Short Answer

Expert verified
So, the solution to the equation \(10^{6-x} = 550\) is \(x = -log_{10} 550 + 6\). This is the value of \(x\) that satisfies the original equation.

Step by step solution

01

Express Equation in Logarithmic Form

The first step is to express the initial equation \(10^{6-x} = 550\) in logarithmic form. This can be achieved by using the formula \(b^y = x \) is equal to \(log_b x = y\). Applying this, the equation can be written as \(log_{10} 550 = 6-x\).
02

Solve for \(x\)

The next step is to solve the equation for \(x\). You can start by subtracting 6 from both sides to isolate the term with \(x\). This results in \(log_{10} 550 - 6 = -x\).
03

Multiply by -1

Since we want \(x\), not \(-x\), the final step is to multiply both sides of the equation by -1. This results in \(-log_{10} 550 + 6 = x\).

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Key Concepts

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

Logarithmic Form
Understanding logarithmic form is essential when working with exponential equations. In essence, a logarithm is another way to express an exponent. The general equation for converting from exponential to logarithmic form is that if we have an equation in the form of \(b^y = x\), then it can be written in logarithmic form as \(\log_b(x) = y\).

Let's apply this principle to the exercise provided. We are given the exponential equation \(10^{6-x} = 550\). To rewrite this equation in logarithmic form, we identify that our base \(b\) is 10, our \(y\) is \(6 - x\), and our \(x\) (in this context, the output of the exponentiation) is 550. Accordingly, we express this as \(\log_{10}(550) = 6 - x\). This reformulation is crucial because it allows us to solve for the variable using properties of logarithms.
Isolating Variables
Isolating the variable is a fundamental step in solving algebraic equations. It involves manipulating the equation so that the variable we are solving for is by itself on one side of the equation. This process typically involves performing inverse operations to 'undo' any operations being applied to the variable.

In our exercise, after converting to logarithmic form, the next task is to isolate \(x\) by getting it on one side of the equation. We start with \(\log_{10}(550) = 6 - x\) and subtract 6 from both sides, leaving us with \(-x = \log_{10}(550) - 6\). From this point, we just need to apply one more operation to solve for \(x\), which is to multiply by -1 to get the variable by itself.
Exponents and Logarithms
The relationship between exponents and logarithms is fundamental in solving exponential equations. Exponents represent repeated multiplication, while logarithms are the inverse operation used to find the exponent that produces a given number.

In our given problem, we're essentially being asked to determine the power to which the base 10 must be raised to yield 550. To make this determination, logarithms are the go-to tool. Knowing that the logarithm is asking 'what power of the base gives me this number?', we've transformed the originally daunting task of solving an exponential equation into a more manageable logarithmic equation. Once in this form and with the variable isolated, reaching the solution is straightforward.

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Most popular questions from this chapter

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