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Fill in the blank to make a true statement. $$ 6^{\square}=\frac{1}{36} $$

Short Answer

Expert verified
The missing exponent is -2.

Step by step solution

01

Understand the Equation

Given the equation: \(6^{\boxed{\text{}}}=\frac{1}{36}\), we need to find the exponent that makes the statement true.
02

Rewrite the Fraction as a Negative Exponent

Recall that a fraction can be rewritten using negative exponents. Since \(6^2 = 36\), \(\frac{1}{36}\) can be written as \(6^{-2}\).
03

Form the Exponent Equality

The equation becomes \(6^{\boxed{\text{}}} = 6^{-2}\).
04

Determine the Exponent

Since the bases are the same, the exponents must be equal. Thus, \(\boxed{\text{}} = -2\).

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Key Concepts

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

Negative Exponents
Negative exponents might seem tricky at first, but they're actually quite simple once you understand the concept. When you see a negative exponent, such as in the expression \(a^{-n}\), it means you take the reciprocal of the base raised to the opposite positive exponent. For example: \[2^{-3} = \frac{1}{2^3} = \frac{1}{8}\] Here, instead of multiplying 2 by itself 3 times, we take 1 divided by 2 multiplied by itself 3 times. This rule can be applied to any number or variable.
Knowing how to rewrite negative exponents as fractions is especially useful when solving equations that involve exponents. For instance, if we have \(6^{\text{unknown}} = \frac{1}{36}\), rewriting the fraction as \(6^{-2}\) allows us to easily determine that the unknown exponent must be -2.
Understanding these properties helps simplify more complex expressions and equations.
Fractional Exponents
Fractional exponents indicate roots. For example, \(a^{\frac{1}{n}}\) represents the n-th root of \(a\). This can be written as: \[8^{\frac{1}{3}} = \text{the cube root of} \, 8 = 2\]
Fractional exponents can also involve numerators other than 1. For instance, \(a^{\frac{m}{n}}\) implies taking the n-th root of \(a\) and then raising it to the m-th power: \[27^{\frac{2}{3}} = (\text{cube root of} \, 27)^2 = 3^2 = 9\]
When dealing with equations that involve fractional exponents, converting them to radicals (roots) is often helpful.
For example, if we need to solve an equation like \(x^{\frac{3}{2}} = 8\), rewriting this as \((\text{square root of} \, x)^3 = 8\) can make it easier to see that \(x=4\), because \(2^3=8\).
Equation Solving
Solving equations with exponents requires understanding and applying the properties of exponents systematically. When faced with an equation such as \(6^{\boxed{\text{}}} = \frac{1}{36}\), you can follow these steps:
  • Identify the expression's current form and consider how exponent properties apply.
  • Rewrite using known properties: For \(6^2 = 36\), the reciprocal \(\frac{1}{36}\) can be written as \(6^{-2}\).
  • Equate the exponents if both sides of the equation have the same base: thus, we know that the exponent we seek is -2.
Solving more complex exponential equations might require additional steps, such as isolating the variable or using logarithms.
The key is always to reduce the problem to a simpler form where properties of exponents can be directly applied.
Practice these steps with various equations to build your confidence and ability to solve them quickly and correctly.

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Most popular questions from this chapter

Graph the following functions on the window [-3,3,1] by [-1,8,1] and comment on the behavior of the graphs near $$ \begin{array}{l} x=0 \\ \mathrm{Y}_{1}=e^{x} \\ \mathrm{Y}_{2}=1+x+\frac{x^{2}}{2} \\ \mathrm{Y}_{3}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6} \end{array} $$

A function of the form \(P(t)=a b^{t}\) represents the population of the given country \(t\) years after January 1,2000 . a. Write an equivalent function using base \(e\); that is, write a function of the form \(P(t)=P_{0} e^{k t} .\) Also, determine the population of each country for the year 2000 . $$\begin{array}{|l|c|c|c|} \hline \text { Country } & P(t)=a b^{t} & P(t)=P_{0} e^{k t} & \begin{array}{c} \text { Population } \\ \text { in } 2000 \end{array} \\ \hline \text { Haiti } & P(t)=8.5(1.0158)^{t} & & \\ \hline \text { Sweden } & P(t)=9.0(1.0048)^{t} & & \\ \hline \end{array}$$ b. The population of the two given countries is very close for the year 2000 , but their growth rates are different. Determine the year during which the population of each country will reach 10.5 million. c. Haiti had fewer people in the year 2000 than Sweden. Why did Haiti reach a population of 10.5 million sooner?

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{2} 15 $$

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\ln x+\ln (x-4)=\ln (3 x-10)\)

Prove the power property of logarithms: \(\log _{b} x^{p}=p \log _{b} x\)

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