Problem 14
Question
Write the exponential equation in logarithmic form. For example, the logarithmic form of \(2^{3}=8\) is log \(_{2} 8=3\) \(24^{0}=1\)
Step-by-Step Solution
Verified Answer
The logarithmic form of \(24^{0}=1\) is log \(_{24} 1 = 0\)
1Step 1: Identify the variables
We identify the base, exponent and value in \(24^{0}=1\). The base \(b\) is 24, the exponent \(x\) is 0 and the output \(y\) equals 1.
2Step 2: Apply the Logarithmic form
We apply the logarithmic form which is log \(_{b} y = x\). Considering the identified variables, we apply this to our equation and get log \(_{24} 1 = 0\)
Key Concepts
Exponential EquationsLogarithm ConversionBase and Exponent Identification
Exponential Equations
Exponential equations are mathematical expressions where a constant base is raised to a variable exponent. For example, in the equation \(2^3 = 8\), the number 2 is raised to the power of 3, resulting in 8. Exponential equations are key in representing growth processes, such as populations or interest rates, where amounts double over fixed intervals.
In math, the general form of an exponential equation is:\[ b^x = y \]
where:
In math, the general form of an exponential equation is:\[ b^x = y \]
where:
- \(b\) is the base
- \(x\) is the exponent
- \(y\) is the result of the base raised to the power of the exponent
Logarithm Conversion
Logarithmic conversion is a vital mathematical process that allows us to express exponential equations in terms of logarithms. Remember that a logarithm is the inverse of an exponential function. So, if you have an exponential equation like \(b^x = y\), the equivalent logarithmic form would be \( \log_{b} y = x \).
This conversion reverses the operations of the original equation, allowing us to solve for encumbered values or verify results more conveniently. Converting exponential equations to logarithms is especially useful in solving problems where you need to find out the exponent, like in our exercise with \(24^0 = 1\). When rewritten in logarithmic form, it becomes \( \log_{24} 1 = 0 \), making it clear and straightforward that the exponent must be 0 to get a result of 1.
This conversion reverses the operations of the original equation, allowing us to solve for encumbered values or verify results more conveniently. Converting exponential equations to logarithms is especially useful in solving problems where you need to find out the exponent, like in our exercise with \(24^0 = 1\). When rewritten in logarithmic form, it becomes \( \log_{24} 1 = 0 \), making it clear and straightforward that the exponent must be 0 to get a result of 1.
Base and Exponent Identification
In any exponential equation, the first step is to clearly identify the different components: the base and the exponent. These form the backbone of both exponential and logarithmic equations.
For instance, in \(24^0 = 1\):
For instance, in \(24^0 = 1\):
- Base \(b\): This is the number that is being repeatedly multiplied. In our example, the base is 24.
- Exponent \(x\): This shows how many times the base is used in multiplication. In this instance, it’s 0, leading to a special case where any number (except 0) raised to the power of 0 equals 1.
- Result \(y\): This is the quantity resulting from raising the base to the exponent. Here, it is 1.
Other exercises in this chapter
Problem 14
Using the Change-of-Base Formula In Exercises \(11-14,\) evaluate the logarithm using the change-of-base formula. Round your result to three decimal places. $$\
View solution Problem 14
Solve for \(x.\) \(\log _{5} x=\frac{1}{2}\)
View solution Problem 15
Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\log _{4} 8$$
View solution Problem 15
Approximate the point of intersection of the graphs of \(f\) and \(g .\) Then solve the equation \(f(x)=g(x)\) algebraically to verify your approximation. \(f(x
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