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Solve the exponential equation algebraically. Approximate the result to three decimal places.\(4^{-t / 3}=0.15\)

Short Answer

Expert verified
The solution to the equation \(4^{-t / 3}=0.15\) is \(t \approx 6.048\).

Step by step solution

01

Rewrite the exponent

We first need to rewrite the equation as \(-t / 3 = \log_4 0.15\). This simplifies the exponential equation making it easier to deal with.
02

Isolate the variable

From the first step equation, isolate variable t by multiplying every term by -3. This gives \(t = -3 \log_4 0.15\).
03

Change of base formula

Recalling logarithm properties, particularly the change of base formula, we can change the base from 4 to 10 (common logarithm). This gives \(t = -3 \left( \frac{\log 0.15}{\log 4} \right)\).
04

Compute using a calculator

Finally, use a calculator to approximate the value of t to three decimal places.

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Key Concepts

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

Logarithm Properties
Understanding logarithm properties is key to solving exponential equations efficiently. Logarithms help in dealing with exponents by transforming them into more manageable forms. Some important properties include:
  • Product Rule: This allows for the splitting up of the logarithm of a product into the sum of logarithms: \( \log_b(xy) = \log_b(x) + \log_b(y) \).

  • Quotient Rule: This rule helps when you have division inside the logarithm: \( \log_b \left( \frac{x}{y} \right) = \log_b(x) - \log_b(y) \).

  • Power Rule: This states that the logarithm of a power can be brought out front: \( \log_b(x^n) = n \cdot \log_b(x) \).

In the exercise, we use the property \( a^{\log_a x} = x \) to rewrite the exponential equation in a logarithmic form. This transformation simplifies the process of isolating the variable.
Change of Base Formula
The change of base formula is a practical tool in logarithms when computation with a specific base doesn't suit our needs. It allows us to convert a logarithm of any base into any other base, typically converting to base 10 or base \( e \), since these are widely supported in calculators.

The formula is expressed as: \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \), where \( k \) is the new base you want to use.

In our problem, we initially have a logarithm with base 4. However, since calculators generally offer functions for base 10 (common logarithm) or base \( e \) (natural logarithm), applying the change of base formula is crucial. By doing this, we can find the needed logarithm using a calculator, leading us to \( t = -3 \left( \frac{\log 0.15}{\log 4} \right) \). This makes the computation feasible and efficient with available tools.
Algebraic Manipulation
Algebraic manipulation involves using algebraic rules to rearrange equations. It’s an essential skill in solving any mathematical problem, including exponential equations.

In our exercise, we have the equation \( -t / 3 = \log_4 0.15 \). To solve for \( t \), we need to isolate it by multiplying each term in the equation by -3, giving us \( t = -3 \cdot \log_4 0.15 \).

This technique involves fundamental algebraic principles: ensuring the variable is alone on one side of the equation, and remembering to perform the same operation on every term to keep the equation balanced. Such steps are crucial for accurately solving equations and finding meaningful results.

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Most popular questions from this chapter

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\frac{3000}{2+e^{2 x}}=2\)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\left(1+\frac{0.0825}{26}\right)^{26 t}=9\)

Domestic Demand The domestic demands \(D\) (in thousands of barrels) for refined oil products in the United States from 1995 to 2005 are shown in the table. (Source: U.S. Energy Information Administration)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 1995 & 6,469,625 \\ \hline 1996 & 6,701,094 \\ \hline 1997 & 6,796,300 \\ \hline 1998 & 6,904,705 \\ \hline 1999 & 7,124,435 \\ \hline 2000 & 7,210,566 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 2001 & 7,171,885 \\ \hline 2002 & 7,212,765 \\ \hline 2003 & 7,312,410 \\ \hline 2004 & 7,587,546 \\ \hline 2005 & 7,539,440 \\ \hline \end{array} $$(a) Use a spreadsheet software program to create a scatter plot of the data. Let \(t\) represent the year, with \(t=5\) corresponding to 1995 . (b) Use the regression feature of a spreadsheet software program to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use the regression feature of a spreadsheet software program to find a logarithmic model \((y=a+b \ln x)\) for the data. (d) Use a spreadsheet software program to graph the exponential model in base \(e\) and the logarithmic model with the scatter plot. (e) Use both models to predict domestic demands in 2008 , 2009, and \(2010 .\) Do both models give reasonable predictions? Explain.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-3 e^{x}-4=0\)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-9 e^{x}-36=0\)

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