Problem 26
Question
Solving an Equation In Exercises \(25-34,\) solve the equation accurate to three decimal places. $$ 5^{6 x}=8320 $$
Step-by-Step Solution
Verified Answer
The value of \( x \) accurate to three decimal places is \(\frac{\ln 8320}{6 \ln 5}\).
1Step 1: Apply Natural Logarithm on both sides
Applying natural logarithm (ln) on both sides of the equation will allow to separate the variable \( x \) from the base. The equation becomes: \n \( \ln (5^{6x}) = \ln (8320) \).
2Step 2: Use the power rule of logarithms
The power rule of logarithms states that for any number \( x \), \( \ln (a^x) = x \ln (a) \). We apply this rule and simplify the equation: \n \( 6x \ln 5 = \ln 8320 \).
3Step 3: Solve for \( x \)
Now we solve for \( x \) in the equation by one more simplification: \n \( x = \frac{\ln 8320}{6 \ln 5} \).
4Step 4: Compute the value of \( x \)
Substituting the natural logarithm values, we find the accurate value up to three decimal places. Calculate and round up to the required decimal places.
Key Concepts
Natural LogarithmPower Rule of LogarithmsExponential and Logarithmic EquationsAlgebraic Manipulation
Natural Logarithm
The natural logarithm, denoted as \textbf{ln}, is a fundamental concept in mathematics, particularly in solving exponential equations. It is the logarithm to the base 'e', where 'e' is an irrational and transcendental number approximately equal to 2.71828. When you take the natural logarithm of a number, you're essentially asking, 'To what power do we raise 'e' to get this number?'
For example, if we have an equation such as \(e^x = y\), the natural logarithm allows us to solve for 'x' by applying \(ln\) to both sides resulting in \(ln(e^x) = ln(y)\), which simplifies to \(x = ln(y)\). This function is especially useful because it transforms the multiplicative processes of exponential equations into the additive processes of logarithms, making them much easier to work with and solve.
For example, if we have an equation such as \(e^x = y\), the natural logarithm allows us to solve for 'x' by applying \(ln\) to both sides resulting in \(ln(e^x) = ln(y)\), which simplifies to \(x = ln(y)\). This function is especially useful because it transforms the multiplicative processes of exponential equations into the additive processes of logarithms, making them much easier to work with and solve.
Power Rule of Logarithms
The power rule of logarithms is an essential rule for simplifying logarithmic expressions, particularly when a variable is an exponent. The rule states that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number: \(\ln(a^x) = x \ln(a)\).
This property allows for the extraction of the exponent in an equation involving an exponential function, making the variable more accessible for algebraic manipulation. When you are dealing with a complex equation like \(5^{6x} = 8320\), implementing the power rule converts the equation into \(6x \ln(5) = \ln(8320)\), which then allows for straightforward division to isolate 'x' and solve the equation.
This property allows for the extraction of the exponent in an equation involving an exponential function, making the variable more accessible for algebraic manipulation. When you are dealing with a complex equation like \(5^{6x} = 8320\), implementing the power rule converts the equation into \(6x \ln(5) = \ln(8320)\), which then allows for straightforward division to isolate 'x' and solve the equation.
Exponential and Logarithmic Equations
Exponential and logarithmic equations involve variables located in exponents and base numbers of logarithms, respectively. Solving these types of equations often involves the use of logarithms to isolate the variable. Because exponential functions and logarithms are inverses of each other, logarithms are particularly effective at 'undoing' exponential functions.
When faced with an equation like \(5^{6x} = 8320\), the exponential form can make it difficult to isolate the variable. By applying a logarithm to both sides, you convert the exponential equation into a form where the variable can be easily isolated and solved, as demonstrated in the example provided.
When faced with an equation like \(5^{6x} = 8320\), the exponential form can make it difficult to isolate the variable. By applying a logarithm to both sides, you convert the exponential equation into a form where the variable can be easily isolated and solved, as demonstrated in the example provided.
Algebraic Manipulation
Algebraic manipulation encompasses the various techniques employed to rearrange and simplify equations to bring them to a solvable form. Once the variable is made the subject, further manipulation can include multiplying or dividing by constants, expanding expressions using algebraic identities, or factoring.
In the context of solving \(5^{6x} = 8320\), once we apply the natural logarithm and use the power rule, we reach \(6x \ln(5) = \ln(8320)\). The next step of algebraic manipulation involves dividing both sides by \(6 \ln(5)\) to isolate 'x'. Through such manipulation, we can derive a formula or direct calculation to find the numerical value of the variable, allowing us to obtain \(x = \frac{\ln 8320}{6 \ln 5}\), a much simpler representation leading to the solution.
In the context of solving \(5^{6x} = 8320\), once we apply the natural logarithm and use the power rule, we reach \(6x \ln(5) = \ln(8320)\). The next step of algebraic manipulation involves dividing both sides by \(6 \ln(5)\) to isolate 'x'. Through such manipulation, we can derive a formula or direct calculation to find the numerical value of the variable, allowing us to obtain \(x = \frac{\ln 8320}{6 \ln 5}\), a much simpler representation leading to the solution.
Other exercises in this chapter
Problem 26
In Exercises 23–32, find the derivative of the function. $$ f(x)=\tanh \left(4 x^{2}+3 x\right) $$
View solution Problem 26
Evaluating a Definite Integral In Exercises \(21-32\) evaluate the definite integral. $$ \int_{1}^{4} \frac{1}{x \sqrt{16 x^{2}-5}} d x $$
View solution Problem 26
In Exercises 19–28, use the properties of logarithms to expand the logarithmic expression. $$ \ln \left(3 e^{2}\right) $$
View solution Problem 26
Finding an Indefinite Integral In Exercises \(1-26,\) find the indefinite integral.. $$ \int \frac{x(x-2)}{(x-1)^{3}} d x $$
View solution