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Solve the exponential equation. Round to three decimal places, when needed. $$10^{2 x^{2}+1}-8=4$$

Short Answer

Expert verified
The solutions of the given equation are \(x=\sqrt{\frac{\log 12 - 1}{2}}\) and \(x=-\sqrt{\frac{\log 12 - 1}{2}}\).

Step by step solution

01

Simplify the Equation

Our given equation is \(10^{2x^2 +1} -8 = 4\). First, we need to simplify it to the form \(10^{2x^2 + 1} = 12\). This allows us to isolate the term containing the variable \(x\).
02

Apply Logarithmic Conversion

Next, we need to transform the equation from its exponential form to a more manageable form using logarithms. For that, we apply the logarithm to both sides of the equation, which gives us \(\log (10^{2x^2 + 1}) = \log 12\). According to the logarithmic rule, this simplifies to \((2x^2 + 1) \log 10 = \log 12\). Since \(\log 10 = 1\), it simplifies further to \(2x^2 + 1 = \log 12\).
03

Solve for X

The equation has been simplified to \(2x^2 + 1 = \log 12\). Arrange the equation to isolate \(x^2 : 2x^2 = \log 12 - 1\). Further simplification gives us \( x^2 = \frac{\log 12 - 1}{2}\). Solving this, we get two solutions for \(x\) owing to the square root, \(x=\sqrt{\frac{\log 12 - 1}{2}}\) and \(x=-\sqrt{\frac{\log 12 - 1}{2}}\).

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Key Concepts

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

Logarithmic Conversion
When dealing with exponential equations, logarithmic conversion acts as a powerful tool to simplify and solve these equations. In essence, a logarithm helps us express the power or exponent needed to obtain a specific value. For example, in our exercise, we’re given the equation: \[10^{2x^2 + 1} = 12\] By converting this equation to its logarithmic form, we can more easily handle and solve for the unknown variable. To achieve this, we apply the logarithm to both sides of the equation. This means that our equation becomes: \[\log (10^{2x^2 + 1}) = \log 12\] Here, we have employed the logarithmic identity: \(\log (a^b) = b \log a\). This formula allows us to "bring down" the exponent in front of the logarithm. It simplifies our original equation to: \[(2x^2 + 1) \log 10 = \log 12\] Since we know \(\log 10 = 1\) (because 10 is the base of the common logarithmic system), the equation reduces to: \[2x^2 + 1 = \log 12\] This straightforward conversion from exponential to logarithmic form makes it much easier to isolate and solve for the variable involved.
Simplifying Equations
Simplifying an equation involves reducing it to its most basic form while retaining its original meaning. In our exercise, the goal is to make the exponential equation easy to handle by detaching any additional constants. Initially, we have the complex equation: \[10^{2x^2 + 1} - 8 = 4\] The first step in simplification is to clear the constant term, '-8', from the left side of the equation. We achieve this by adding 8 to both sides, which results in: \[10^{2x^2 + 1} = 12\] This step is crucial because it allows us to focus only on the term containing the unknown variable, which is a necessary process for the subsequent application of logarithmic conversion. By isolating the exponential part of the equation, we create simpler terms that can be solved more intuitively. Upon achieving this form, we have a streamlined path to solving for our variable \(x\).
Solving for Variables
Solving for variables is the final step in resolving an equation where we need to find the value of the unknown quantities. After converting and simplifying our logarithmic equation, we operate within the form: \[2x^2 + 1 = \log 12\] From here, we aim to isolate \(x^2\) by rearranging the equation: \[2x^2 = \log 12 - 1\] And then simply dividing all terms by 2: \[x^2 = \frac{\log 12 - 1}{2}\] To find the possible values of \(x\), the square root of both sides is taken. It's important to remember that squaring introduces two solutions:
  • \(x = \sqrt{\frac{\log 12 - 1}{2}}\)
  • \(x = -\sqrt{\frac{\log 12 - 1}{2}}\)
These solutions encompass both the positive and negative square root possibilities. By solving for \(x\) in this way, you determine all potential values for \(x\) that satisfy the original equation.

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Most popular questions from this chapter

The spread of a disease can be modcled by a logistic function. For example, in carly 2003 there was an outbreak of an illness called SARS (Severe Acute Respiratory Syndrome) in many parts of the world. The following table gives the total momber of cases in Canada for the wecks following March 20,2003 (Source: World Health Organization) (Note: The total number of cases dropped from 149 to 140 between weeks 3 and 4 because some of the cases thought to be SARS were reclassified as other discases.) $$\begin{array}{|c|c|}\hline\text { Weeks since } & \\\\\text { March } 20,2003 & \text { Total Cases } \\\0 & 9 \\\1 & 62 \\\2 & 132 \\\3 & 149 \\\4 & 140 \\\5 & 216 \\\6 & 245 \\\7 & 252 \\\8 & 250 \\\\\hline\end{array}$$ (a) Explain why a logistic function would suit this data well. (b) Make a scatter plot of the data and find the logistic function of the form \(f(x)=\frac{\epsilon}{1+a \varepsilon^{-1}}\) that best fits the data. (c) What docs \(c\) signify in your model? (d) The World Health Organization declared in July 2003 that SARS no longer posed a threat in Canada. By analyring this data, explain why that would be so.

Evaluate the expression to four decimal places using a calculator. $$\log 2500$$

A new car that costs $$\$ 25,000$$ depreciates to $$80 \%$$ of its value in 3 years. (a) Assume the depreciation is linear. What is the linear function that models the value of this car \(t\) years after purchase? (b) Assume the value of the car is given by an exponential function \(y=A e^{h t},\) where \(A\) is the initial price of the car. Find the value of the constant \(k\) and the exponential function. (c) Using the linear model found in part (a), find the value of the car 5 years after purchase. Do the same using the exponential model found in part (b). (d) Graph both models using a graphing utility. Which model do you think is more realistic, and why?

In \(1965,\) Gordon Moore, then director of Intel research, conjectured that the number of transistors that fit on a computer chip doubles every few years. This has come to be known as Moore's Law. Analysis of data from Intel Corporation yields the following model of the number of transistors per chip over time: $$s(t)=2297.1 e^{0.3316 t}$$ where \(s(t)\) is the number of transistors per chip and \(t\) is the number of years since \(1971 .\) (Source: Intel Corporation) (a) According to this model, what was the number of transistors per chip in \(1971 ?\) (b) How long did it take for the number of transistors to double?

This set of exercises will draw on the ideas presented in this section and your general math background. What is wrong with the following step? $$\log x+\log (x+1)=0 \Rightarrow x(x+1)=0$$

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