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Chapter 3: Problem 14

Solve for \(x\). $$\log _{5} x=\frac{1}{2}$$

Short Answer

Expert verified
The solution to the given equation is \(x = \sqrt{5}\)

Step by step solution

01

Understanding the logarithmic format

In a logarithmic equation \(\log_{b}a = c\), the base \(b\) raised to the power of \(c\) equals \(a\). We can convert the given equation to this format.
02

Convert the logarithmic equation to an exponential equation

By reformatting \(\log_{5}x=\frac{1}{2}\) into the exponential form, we get \(5^{\frac{1}{2}} = x\)
03

Simplifying the exponential equation

The equation \(5^{\frac{1}{2}} = x\) simplifies to \(\sqrt{5} = x\)

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Key Concepts

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

Logarithmic to Exponential Form
Understanding how to translate a logarithmic equation into exponential form is vital to solving logarithmic equations. Let's break this down with the example problem, where we have \(\log_{5} x = \frac{1}{2}\).

In general, the logarithmic equation \(\log_{b}a = c\) implies that the base \(b\) raised to the exponent \(c\) equals \(a\). Therefore, to convert our problem from logarithmic to exponential form, we rewrite \(\log_{5}x\) as \(5^{\frac{1}{2}} = x\). This is a critical step, as it transforms the problem into an equation involving exponentiation, which is often more intuitive for students to solve.
Exponents and Radicals
When we deal with exponents and radicals, we're essentially looking at different ways to express the same mathematical concepts. In our example, \(5^{\frac{1}{2}} = x\), the exponent \(\frac{1}{2}\) indicates a square root.

To understand this, remember that an exponent of \(\frac{1}{n}\) means the nth root of a number. For instance, \(a^{\frac{1}{2}}\) is the square root of \(a\), and \(a^{\frac{1}{3}}\) is the cube root. So, \(5^{\frac{1}{2}}\) simplifies to \(\sqrt{5}\), which means that \(x\) is the square root of \(5\). Recognizing these relationships is key to simplifying and solving problems involving exponents and radicals.
Logarithm Properties
Logarithms have specific properties that make solving equations efficient and straightforward. One of the fundamental properties is the one-to-one correspondence between logarithms and exponents, used in the first concept. Other properties include:

  • Product Rule: \(\log_{b}(mn) = \log_{b}(m) + \log_{b}(n)\), which states that the logarithm of a product is the sum of the logarithms.
  • Quotient Rule: \(\log_{b}\frac{m}{n} = \log_{b}(m) - \log_{b}(n)\), meaning the logarithm of a quotient is the difference of the logarithms.
  • Power Rule: \(\log_{b}(m^n) = n\log_{b}(m)\), which allows us to take an exponent out of the logarithm.

All these properties can be applied to simplify and solve logarithmic equations efficiently. For example, applying the Power Rule can simplify the process of dealing with exponents within a logarithm.

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Most popular questions from this chapter

In Exercises \(103-106,\) use the change-of-base formula to rewrite the logarithm as a ratio of logarithms. Then use a graphing utility to graph the ratio. $$f(x)=\log _{1 / 2} x$$

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$5^{x}=212$$

Population The time \(t\) (in years) for the world population to double when it is increasing at a continuous rate of \(r\) is given by \(t=(\ln 2) / r\) (a) Complete the table and interpret your results. \begin{tabular}{|l|l|l|l|l|l|l|}\hline\(r\) & 0.005 & 0.010 & 0.015 & 0.020 & 0.025 & 0.030 \\\\\hline\(t\) & & & && &..\begin{array}{|l|l|l|l|l|l|l|} \hline r & 0.005 & 0.010 & 0.015 & 0.020 & 0.025 & 0.030 \\ \hline t & & & & & & \\ \hline \end{array}

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x^{2} e^{2 x}+2 x e^{2 x}=0$$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log _{2} x+\log _{2}(x+2)=\log _{2}(x+6)$$
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