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Chapter 11: Problem 6

Rewrite the equation using exponents instead of logarithms. $$ \log \frac{1}{10^{3}}=-3 $$

Short Answer

Expert verified
Question: Rewrite the logarithmic equation \(\log{\frac{1}{10^{3}}}=-3\) using exponents. Answer: \(10^{-3}=\frac{1}{10^3}\)

Step by step solution

01

Rewrite the logarithm as an exponent

Using the property of logarithms, rewrite the equation \(\log{\frac{1}{10^{3}}}=-3\) as \(10^{-3}=\frac{1}{10^3}\).
02

Simplify the right-hand side

We know that \(10^{-3}\) is the reciprocal of \(10^3\). So, we can rewrite the equation as: \(10^{-3}=\frac{1}{10^3}\).

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Key Concepts

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

Logarithms
Logarithms are fundamentally the inverse operation of exponentiation. When you are given a logarithmic expression such as \(\log_{b}(x) = y\), it means that the base \(b\) raised to the power of \(y\) equals \(x\). This is the same as saying \(b^y = x\). For example, if you see \(\log_{10}(1000)\), you're finding the power to which the number 10 must be raised to get 1000. In this case, it's 3 because 10 to the power of 3 is 1000.
It's essential to remember that logs can be expressed in terms of exponents. This is especially helpful for converting and solving logarithmic equations.
Equation rewriting
Rewriting equations is a powerful technique, especially when dealing with logarithms. The aim is to express a logarithmic equation in simpler terms using exponent laws.
Let's take the example \(\log{\left( \frac{1}{10^3} \right) = -3 }\). This tells us that the base 10 raised to the power of \(-3\) equals \(\frac{1}{10^3}\). Rewriting this gives us the equation \(10^{-3} = \frac{1}{10^3}\).
Why rewrite? Because it simplifies the problem and sometimes reveals the solution immediately. Once in a different form, these equations can become more manageable or insightful.
Reciprocal
The concept of reciprocals is derived from fractions and multiplicative inverses. A reciprocal is simply one of a pair of numbers that, when multiplied together, equals 1. For any non-zero number \(x\), its reciprocal is \(\frac{1}{x}\).
When you have an expression like \(10^{-3}\), it's a perfect example of reciprocals in action because it equals \(\frac{1}{10^3}\). You can think of the negative exponent as a signal to find the reciprocal of the base raised to the positive power. Understanding this helps to simplify exponents when working with logarithms.
  • Reciprocals turn divisions into multiplications.
  • They help solve or simplify equations, especially involving powers and roots.
Embracing this concept can make dealing with negative exponents much more intuitive.

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Most popular questions from this chapter

Rewrite the expression in terms of \(\log A\) and \(\log B\), or state that this is not possible. $$ \log (1 /(A B)) $$

Write the expressions in Problems \(44-49\) in the form \(\log _{b} x\) for the given value of \(b\). State the value of \(x\), and verify your answer using a calculator. $$ \frac{\log 17}{2}, \quad b=100 $$

If possible, use logarithm properties to rewrite the expressions in terms of \(u, v, w\) given that $$u=\log x, v=\log y, w=\log z$$ Your answers should not involve logs. $$ \left(\log \frac{1}{y^{3}}\right)^{2} $$

Iodine-131, used in medicine, has a half-life of 8 days. (a) If \(5 \mathrm{mg}\) are stored for a week, how much is left? (b) How many days does it take before only \(1 \mathrm{mg}\) remains?

Assume \(a\) and \(b\) are positive constants. Imagine solving for \(x\) (but do not actually do so). Will your answer involve logarithms? Explain how you can tell. $$ 10^{2} x=10^{3}+10^{2} $$
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