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Write the exponential equation in logarithmic form. For example, the logarithmic form of \(2^{3}=8\) is \(\log _{2} 8=3\). $$5^{3}=125$$

Short Answer

Expert verified
The logarithmic form of \(5^{3}=125\) is \(\log_{5}125 = 3\).

Step by step solution

01

Recognize the structure of the exponential form and correlate with logarithmic form

The given exponential equation is \(5^{3}=125\), where \(5\) is the base, \(3\) is the exponent, and \(125\) is the result. The structure of the logarithmic form is \(\log _{b} y=x\), where \(b\) is the base, \(y\) is the value we're taking the log of, and \(x\) is the result. As we see, in the exponential form, \(b=5\), \(x=3\) and \(y=125\). These values correlate to the respective positions in the logarithmic form.
02

Apply values to the logarithmic form

Knowing that the base of the logarithm will be the base from the exponential form (\(5\) in our example), we put this as the base of the log. The number we are taking the log of is the result from the exponential form (\(125\) in our example), we place this inside the log. The result of the log will be the exponent from the exponential form (\(3\), in our example).

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Key Concepts

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

Converting Exponents to Logarithms
Understanding how to convert exponents to logarithmic form is essential for solving various algebra problems, as it helps in uncovering the relationship between exponential and logarithmic functions, which are inverses of each other. The primary rule for conversion is to recognize that for any given exponential equation of the form b^{x}=y, where b is the base, x is the exponent, and y is the result, the equivalent logarithmic form is log_b(y)=x.

This conversion is based on the definition of logarithms. Simply put, the logarithm log_b(y) asks the question: 'To what power must the base b be raised, in order to produce y?' The answer to that question, naturally, is the exponent x. Thus, if we have an exponential equation like 5^{3}=125, we can express this as the logarithmic equation log_5(125)=3. Here, the base 5 raised to the power of 3 gives 125.

Let's consider a few examples:
  • In the expression 2^{4}=16, the logarithmic form would be log_2(16)=4.
  • For 10^{2}=100, the logarithmic form is log_{10}(100)=2, which is often written simply as log(100)=2 since base 10 is commonly assumed in logs without a specified base.
Converting exponents to logarithms allows us to solve equations that would otherwise be difficult to tackle. This conversion is a powerful tool in algebra that facilitates the manipulation of equations to find unknown variables.
Logarithmic Equations
Logarithmic equations are equations that contain a logarithm with a variable that you need to solve for. They can often be solved by leveraging the properties of logarithms, such as the product, quotient, and power rules. The key to solving logarithmic equations is to either isolate the logarithm on one side of the equation or convert the logarithmic equation into an exponential form in order to simplify and solve for the variable.

Here are the basic steps to solve a logarithmic equation:
  • Isolate the logarithmic part of the equation if possible.
  • Convert the logarithmic equation into its equivalent exponential form.
  • Solve the resulting equation for the variable.
As an example, consider the logarithmic equation log_3(x)=4. To solve for x, we can convert this to its exponential form: 3^{4}=x, yielding the solution x=81.

Another common technique is to use the property of logarithms that states if log_b(M)=log_b(N), then M=N. This is especially helpful when the equation contains logarithms with like bases on either side. It's important to check for extraneous solutions, which are apparent solutions that don't actually satisfy the original equation, as not all values may be valid within the domain of logarithmic functions.
Exponential Equations
Exponential equations are equations in which the variable appears as an exponent. Solving them usually involves isolating the term with the variable and applying logarithms to both sides of the equation to simplify and solve it. The process often starts by expressing both sides of the equation with the same base if possible, and then using the property of exponents that if a^x=a^y, then x=y, given that a is positive and not equal to 1.

Some common steps to solve exponential equations are:
  • Expressing all terms with the same base if possible.
  • Equating the exponents of these terms.
  • Solving the resulting equation for the variable.
If an equation cannot be easily rewritten with a common base, applying natural logarithms (ln) to each side can be useful due to the property that ln(a^x)=x*ln(a). For example, to solve the equation 2^x=32, we can rewrite 32 as 2^5, which gives us 2^x=2^5. Equating the exponents, we find that x=5.

For cases that are less straightforward, such as 2^x=3, a natural logarithm can be applied to both sides: ln(2^x)=ln(3), simplifying to x*ln(2)=ln(3), and solving for x yields x=ln(3)/ln(2). Understanding exponential equations is critical in various disciplines, including finance, science, and technology, as they often model phenomena involving growth or decay.

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Most popular questions from this chapter

Use a graphing utility to graph the functions \(y_{1}=\ln x-\ln (x-3)\) and \(y_{2}=\ln \frac{x}{x-3}\) in the same viewing window. Does the graphing utility show the functions with the same domain? If so, should it? Explain your reasoning.

A cup of water at an initial temperature of \(78^{\circ} \mathrm{C}\) is placed in a room at a constant temperature of \(21^{\circ} \mathrm{C}\). The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form \((t, T),\) where \(t\) is the time (in minutes) and \(T\) is the temperature (in degrees Celsius). $$\begin{aligned} &\left(0,78.0^{\circ}\right),\left(5,66.0^{\circ}\right),\left(10,57.5^{\circ}\right),\left(15,51.2^{\circ}\right)\\\ &\left(20,46.3^{\circ}\right),\left(25,42.4^{\circ}\right),\left(30,39.6^{\circ}\right) \end{aligned}$$ (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \((t, T)\) and \((t, T-21)\). (b) An exponential model for the data \((t, T-21)\) is given by \(T-21=54.4(0.964)^{t} .\) Solve for \(T\) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use the graphing utility to plot the points \((t, \ln (T-21))\) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \(\ln (T-21)=a t+b\) Solve for \(T,\) and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \(y\) -coordinates of the revised data points to generate the points $$\left(t, \frac{1}{T-21}\right)$$. Use the graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. The resulting line has the form $$\frac{1}{T-21}=a t+b$$. Solve for \(T,\) and use the graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562, \log _{b} 3 \approx 0.5646,\) and \(\log _{b} 5 \approx 0.8271.\) $$\log _{b} \sqrt{2}$$

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

Condense the expression to the logarithm of a single quantity. $$\frac{2}{3} \log _{7}(z-2)$$

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