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Chapter 3: Problem 34

Find the exact value of the logarithmic expression without using a calculator. (If this is not possible, state the reason.) $$\log _{3} 81^{-3}$$

Short Answer

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Step by step solution

01

Express number inside the logarithm in terms of base

Rewrite 81 as a power of 3 because 3 is the base of the logarithm in the expression. So, \[81 = 3^4\].
02

Apply the power rule

Substitute \(3^4\) into the expression and reinforce the exponent to -3: \[\log _{3} (3 ^ 4) ^ {-3}\]. Now, apply the power rule of logarithms, which states that \[log_b(a^n)=n * log_b(a)\], where n=-3 and a = \(3^4\). This gives us \[-3*\log _{3} (3 ^ 4)\].
03

Simplify the logarithmic term

Simplify the term \[\log_{3}(3^4)\]. This represents the power we must raise 3 to obtain \(3^4\), which is 4. Substituting this equation gives \[-3*4\].
04

Simplify the expression

Lastly, simply multiply the constants to find the answer to the logarithmic expression: \[-3*4=-12\].

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Key Concepts

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

Logarithmic Expression
Understanding a logarithmic expression is crucial in solving any logarithmic equation. A logarithm answers the question: "To what power must the base be raised, in order to get a certain number?" For example, in the expression \(\log_{b}(x)\), where \(b\) is the base and \(x\) is the number, you are finding the power to which \(b\) must be raised to yield \(x\). In the given exercise, \(\log_{3} 81^{-3}\), the base \(b\) is 3, and the number is \(81^{-3}\).
  • The goal is to express \(81^{-3}\) in a form that makes use of the base 3.
  • This simplifies the process of solving the expression.
Power Rule of Logarithms
The power rule of logarithms is a handy tool for breaking down expressions involving exponents. It states that \(\log_b(a^n) = n \cdot \log_b(a)\). Essentially, you can "bring down" the exponent \(n\) as a multiplier outside the logarithmic term.
This rule simplifies calculations since it transforms the problem into a basic multiplication task.
  • When applying this to \(\log_3(81^{-3})\), start by rewriting 81 as \(3^4\).
  • This changes the expression to \(\log_3((3^4)^{-3})\).
  • Apply the power rule to get \(-3 \cdot \log_3(3^4)\).

This intermediate step helps to make further simplifications easy.
Simplifying Logarithms
Simplifying logarithms often involves recognizing the base and the power that give a direct simplification. Once a logarithmic expression is rewritten using the base, simplifying becomes much easier.
In our example, the simplification step uses the expression \(\log_3(3^4)\). Here, the base 3 raised to what power equals \(3^4\)?
  • The answer is simply 4 because \(3^4\) is already in the form of the power associated with base 3.
  • Thus, \(\log_3(3^4) = 4\).

Simplification allows the problem to reduce to a multiplication task, \(-3 \times 4\), leading to an easier solution.
Exact Value Without Calculator
Finding the exact value of a logarithmic expression without a calculator involves a deep understanding of how logarithms work. Calculations are based on properties and rules of logarithms rather than digitized computations.
With the simplified expression \(-3 \times \log_3(3^4)\), where \(\log_3(3^4) = 4\), multiply \(-3\) and 4.
  • This results in \(-3 \times 4 = -12\).
  • Hence, the exact value of \(\log_3(81^{-3})\) is \(-12\).
  • Importantly, the power and base relation leads us directly to this accurate outcome.

With these steps, you unearth the precise value ensured by logical operations rather than computations on a calculator.

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Most popular questions from this chapter

A $$\$ 120,000$$ home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of $$\$ 839.06 .$$ Part of the monthly payment is paid toward the interest charge on the unpaid balance, and the remainder of the payment is used to reduce the principal. The amount that is paid toward the interest is \(u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) and the amount that is paid toward the reduction of the principal is \(v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the larger part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?

COMPARING MODELS If $$\$ 1$$ is invested in an account over a 10 -year period, the amount in the account, where \(t\) represents the time in years, is given by \(A=1+0.06 \llbracket t \rrbracket\) or \(A=[1+(0.055 / 365)]^{[365 t]}\) depending on whether the account pays simple interest at \(6 \%\) or compound interest at \(5 \frac{1}{2} \%\) compounded daily. Use a graphing utility to graph each function in the same viewing window. Which grows at a higher rate?

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\ln x+\ln (x-2)=1$$

The total interest \(u\) paid on a home mortgage of \(P\) dollars at interest rate \(r\) for \(t\) years is \(u=P\left[\frac{r t}{1-\left(\frac{1}{1+r / 12}\right)^{12 t}}-1\right]\) Consider a $$\$ 120,000$$ home mortgage at \(7 \frac{1}{2} \%\). (a) Use a graphing utility to graph the total interest function. (b) Approximate the length of the mortgage for which the total interest paid is the same as the size of the mortgage. Is it possible that some people are paying twice as much in interest charges as the size of the mortgage?

Use the Richter scale \(R=\log \frac{l}{I_{0}}\) for measuring the magnitudes of earthquakes. Find the magnitude \(R\) of each earthquake of intensity \(I\) (let \(I_{0}=1\) ). (a) \(I=199,500,000\) (b) \(I=48,275,000\) (c) \(I=17,000\)
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