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Solve for \(x\).\(7^{x}=\frac{1}{49}\)

Short Answer

Expert verified
The solution to the equation is \(x = -2\).

Step by step solution

01

Rewrite the fraction

The first step is to rewrite the fraction \(\frac{1}{49}\) as \(7^{-2}\) because \(\frac{1}{49}\) equals the same value as \(7^{-2}\).
02

Set the exponents equal to each other

Next, set the exponents in the equation equal to each other. In this case, \(x\) and \(-2\) are the exponents. So a new equation \(x = -2\) can be formed.
03

State the solution for \(x\)

The solution for \(x\) has already been obtained. So, check if the solution works in the original equation by substituting the value of \(x\) back into the equation. This will confirm whether the solution fits in the equation or not.

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Key Concepts

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

Exponents
Exponents are a mathematical shorthand to express repeated multiplication of the same number, known as the base, raised to a power, which represents the number of times to multiply the base by itself. In the equation \(7^x = \frac{1}{49}\), '7' is the base and 'x' is the exponent. Understanding exponents is crucial because they are prevalent in various fields including science, finance, and engineering.

A negative exponent, such as \(-2\), indicates that the base should be taken as the reciprocal and multiplied by itself a positive number of times. For instance, \(7^{-2}\) corresponds to \(\frac{1}{7^2}\), which is \(\frac{1}{49}\). When solving exponential equations, we want to represent both sides of the equation with the same base and exponent if possible, just like the step-by-step solution we see with \(7^x = 7^{-2}\), facilitating a simplified solution.
Inverse Operations
Inverse operations are mathematical operations that reverse the effect of another operation. They are foundational for solving equations. For example, addition and subtraction are inverse operations, just as multiplication and division are. Exponents have their own inverse operation called roots.

When solving the equation \(7^x = \frac{1}{49}\), we use the principle of inverse operations to rewrite \(\frac{1}{49}\) as an expression with the same base of the opposite side: \(7^x = 7^{-2}\). By identifying that raising a number to an exponent and taking a root are inverse operations, we can equate the exponents of the same base directly. This leads to the simple algebraic equation \(x = -2\), bypassing the need for more complicated operations.
Logarithmic Functions
Logarithmic functions are the inverses of exponential functions. While exponential functions deal with the question 'The base raised to what power gives me this number?', logarithms ask 'To what power must I raise the base to get this number?'. They are denoted as \(\text{log}_b (x)\), which reads as 'log base \(b\) of \(x\)'.

When solving exponential equations where the bases can't be easily made the same, or when dealing with more complex equations, logarithms become an essential tool. For instance, if the given problem was more complex, such as \(7^x = 20\), the use of logarithms would be necessary. You would take the logarithm of both sides, resulting in \(x = \text{log}_7 (20)\), as a way to isolate \(x\) and solve the equation. Recognizing when and how to apply logarithmic functions is a key skill in managing exponential equations.

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Most popular questions from this chapter

Classify the model as an exponential growth model or an exponential decay model.\(y=4 e^{0.07 t}\)

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

Population The population \(P\) of the United States officially reached 300 million at about 7:46 A.M. E.S.T. on Tuesday, October 17,2006 . The table shows the U.S. populations (in millions) since 1900. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1900 & 76 \\ \hline 1910 & 92 \\ \hline 1920 & 106 \\ \hline 1930 & 123 \\ \hline 1940 & 132 \\ \hline 1950 & 151 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1960 & 179 \\ \hline 1970 & 203 \\ \hline 1980 & 227 \\ \hline 1990 & 250 \\ \hline 2000 & 282 \\ \hline 2006 & 300 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1900 . (b) Use the regression feature of a graphing utility 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) Graph the exponential model in base \(e\) with the scatter plot of the data. What appears to be happening to the relationship between the data points and the regression curve at \(t=100\) and \(t=106 ?\) (d) Use the regression feature of a graphing utility to find a logistic growth model for the data. Graph each model using the window settings shown below. Which model do you think will give more accurate predictions of the population well beyond \(2006 ?\)

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

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{10} 3 z=2\)

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