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Convert to logarithmic form. $$x^{5}=995$$

Short Answer

Expert verified
\(\log_x(995) = 5\)

Step by step solution

01

Understand the Exponential Form

Take note of the exponential form of the equation which is written as: 'base' raised to the power of 'exponent' equals 'result', or in this case, as: \(x^5 = 995\). Here, \(x\) is the base, \(5\) is the exponent, and \(995\) is the result.
02

Apply the Definition of a Logarithm

Recall the definition of a logarithm: if \(b^y = x\), then \(\log_b(x) = y\). We can apply this definition to the given equation \(x^5 = 995\) by setting \(b = x\), \(y = 5\), and \(x = 995\).
03

Write the Logarithmic Form

Using the definition of logarithms, write the equation in logarithmic form: \(\log_x(995) = 5\). This expresses the same relationship as the original equation, just in logarithmic form.

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Key Concepts

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

Exponential to Logarithmic Form Conversion
When we look at equations in exponential form, such as the textbook example of converting the exponential equation \(x^{5}=995\), there is an implicit challenge of understanding the relationship between the base, exponent, and result. This relationship becomes more transparent when we convert the equation into logarithmic form.

In general, if we have an equation \(b^y = x\), this can be expressed in logarithmic form as \(\log_{b}(x) = y\). This rephrasing allows us to see what power \(y\) we must raise \(b\) to, in order to get \(x\).

Using this concept, let's perform the transition from exponential to logarithmic form for our specific example:
  • Identify the base (\(b\)) which is \(x\), the exponent (\(y\)) which happens to be 5, and the result (\(x\)) which is 995.
  • Apply the definition of a logarithm: \(b^y = x\) would translate to \(\log_{b}(x) = y\).
  • Substitute in our values to convert \(x^{5}=995\) into \(\log_{x}(995) = 5\).
Through this method, we can transform any exponential equation into an equivalent logarithmic equation, providing a different perspective on the relationship between numbers in the equation.
Definition of a Logarithm
Understanding the fundamental definition of a logarithm is vital for grappling with various mathematical concepts. A logarithm basically answers the question: To what exponent must we raise a specific base, in order to produce a given number?

In its technical definition, if we have an equation of the form \(b^y = x\), the logarithm of \(x\) with base \(b\) is \(y\), which is symbolically represented as \(\log_{b}(x) = y\). Here \(b\) is the base of the logarithm, \(x\) is the number we want to find the power of \(b\) that will produce, and \(y\) is that power or exponent.

To illustrate, if we have \(2^3 = 8\), then \(\log_{2}(8) = 3\). This tells us that 2 must be raised to the power of 3 to yield 8.

This definition is an essential bridge between the realms of exponential and logarithmic functions, and an understanding of this relationship is crucial in solving equations and understanding growth patterns in real-life scenarios.
Logarithms in Mathematics
Logarithms hold a special place in mathematics for their unique properties and various applications. They are indispensable tools in solving exponential equations, especially when dealing with growth and decay problems, such as interest in finance or population dynamics in biology.

One of the most useful properties of logarithms is the ability to turn multiplication into addition and division into subtraction. This is particularly helpful when working with very large or very small numbers, as it simplifies calculations.

For example, the logarithmic property that states \(\log_{b}(mn) = \log_{b}(m) + \log_{b}(n)\) allows us to add the logarithms of \(m\) and \(n\) rather than multiplying \(m\) and \(n\) directly, which can be a much simpler process.

Furthermore, logarithms are the foundation of logarithmic scales, which are used to measure the intensity of earthquakes (Richter scale), sound (decibels), and the acidity of solutions (pH scale). Their ability to compress wide-ranging quantities into a manageable scale makes them invaluable in many scientific fields.

In advanced mathematics, logarithms extend into complex numbers, contributing to the field of complex analysis. They are also a fundamental component in calculus, assisting in the integration and differentiation of exponential functions. Considering their versatility and utility, mastering logarithms opens doors to understanding many different areas of mathematics and science.

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Most popular questions from this chapter

The infinite series often used to calculate \(e^{x}\) is as follows:$$\begin{array}{|lc|}\hline \text { Series } & \\\\\text { Approximation } & e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots & 159 \\\\\text { for } e^{x} & & \\\\\hline\end{array}$$.Write a program to compute \(e^{x}\) by using the first 15 terms of this series, and use it to find \(e^{5}\).

A certain long pendulum, released from a height \(y_{0}\) above its rest position, will be at a height $$y=y_{0} e^{-0.75 t}$$ at \(t\) seconds. If the pendulum is released at a height of \(15 \mathrm{cm},\) at what time will the height be \(5.0 \mathrm{cm} ?\)

If a resource is being used up at a rate that increases exponentially, the time it takes to exhaust the resource (called the exponential expiration time, EET) is $$\mathrm{EET}=\frac{1}{n} \ln \left(\frac{n R}{r}+1\right)$$ where \(n\) is the rate of increase in consumption, \(R\) the total amount of the resource, and \(r\) the initial rate of consumption. If we assume that the United States has oil reserves of \(207 \times 10^{9}\) barrels and that our present rate of consumption is \(6.00 \times 10^{9}\) barrels/yr, how long will it take to exhaust these reserves if our consumption increases by \(7.00 \%\) per year?

Using the formula for compound interest, Eq. \(1009, y=a(1+n)^{t},\) calculate the number of years it will take a sum of money to triple when invested at a rate of \(12 \%\) per year.

Find the common logarithm of each number. $$4.83$$

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