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Use the definition of a logarithm to write the equation in exponential form. For example, the exponential form of \(\log _{5} 125=3\) is \(5^{3}=125\).\(\log _{2} \frac{1}{2}=-1\)

Short Answer

Expert verified
The exponential form of the equation \(\log _{2} \frac{1}{2}=-1\) is \(2^{-1} = \frac{1}{2}\)

Step by step solution

01

Recognize the log structure

Given is \(\log _{2} \frac{1}{2}= -1\). The base of the logarithm is 2, the output (or 'answer') of the logarithm is -1, and the quantity inside the logarithm is \( \frac{1}{2}\).
02

Use the definition of a logarithm

According to the definition of a logarithm, \(\log_b a = c\) translates to \(b^c = a\). Using this rule, the base 2 is raised to the power of -1 to yield \(\frac{1}{2}\). This way the logarithmic equation can be converted to an exponential one.
03

Write the equation in exponential form

Translated the equation becomes \(2^{-1} = \frac{1}{2}\)

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Key Concepts

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

Logarithm Definition
The concept of a logarithm is fundamental in mathematics, particularly when dealing with exponential relationships. A logarithm, in essence, expresses the power to which a number (known as the base) has to be raised to obtain another number. Mathematically, if we have a base b, and we want to know what power c we must raise b to get a, we write this as \[ \log_b a = c \.\] For instance, in the exercise provided, the logarithm \(\log _{2} \frac{1}{2} = -1\) essentially asks 'To what power must we raise 2 to get \(\frac{1}{2}\)?' The answer is -1 because \(2^{-1} = \frac{1}{2}\).
Understanding logarithms includes recognizing their parts: the base b, the number a (sometimes called the argument of the logarithm), and the result c, which is the power. Remember that logarithms can only be taken of positive numbers, and the base b must always be positive and not equal to 1.
Exponential Equations
Exponential equations are mathematical expressions where the variable appears in the exponent. This type of equation is of the form \[ b^x = a\], where b is the base, x is the exponent, and a is the result. To solve these equations, we often rely on logarithms to 'bring down' the exponent so that we can solve for the variable.
In daily life, exponential equations model many natural phenomena such as compound interest, population growth, and radioactive decay. To understand exponential equations, it is essential to grasp the power of exponential functions, which show a constant rate of growth or decay. For example, in an exponential relationship, a quantity doubles or halves over consistent intervals, exhibiting a straightforward pattern despite the rapid increase or decrease in the actual numbers.
Logarithmic Equations
Logarithmic equations contain logarithms and typically involve finding the value of a variable located inside the logarithm. These equations can be written in the form \[\log_b(x) = c\], where you're often asked to solve for x. Just like the exercise example \(\log _{2} \frac{1}{2} = -1\), logarithmic equations can be tricky because the solution requires transforming the logarithmic form into an exponential form to solve for the variable.
Solving logarithmic equations often involves properties of logarithms such as the product rule, the quotient rule, and the power rule. Using these properties, you can manipulate the equation to isolate the variable and find the solution. For the students, it's paramount to understand how to use the definition of a logarithm to rewrite the equation in an alternative form that reveals solutions more accessible, just as in the step-by-step solution provided where we transition from logarithmic to exponential form. It's also important to remember that solutions must be checked for extraneous values, as not all solutions derived algebraically may satisfy the original equation.

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Most popular questions from this chapter

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(1000 e^{-4 x}=75\)

The value \(y\) (in billions of dollars) of U.S. currency in circulation (outside the U.S. Treasury and not held by banks) from 1996 to 2005 can be approximated by the model \(y=-302+374 \ln t, \quad 6 \leq t \leq 15\) where \(t\) represents the year, with \(t=6\) corresponding to 1996\. (Source: Board of Governors of the Federal Reserve System) (a) Use a graphing utility to graph the model. (b) Use a graphing utility to estimate the year when the value of U.S. currency in circulation exceeded \(\$ 600\) billion. (c) Verify your answer to part (b) algebraically.

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Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0725\)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(-14+3 e^{x}=11\)

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