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Solve the equation. $$e^{x \ln 3}=27$$

Short Answer

Expert verified
x = 3

Step by step solution

01

Recall Logarithm and Exponential Properties

Recall that the expression can be manipulated using the properties of logarithms and exponents. The most relevant property is that if you have an exponential expression with a base of the natural number \( e \, s: e^{a \, ln \, n} = n^a. \), we can rewrite given problem as \( 3^x = 27 \).
02

Express 27 as Power of 3

Express 27 as a power of 3. We know that 27 can be written as \( 3^3 \). Hence, the equation becomes \( 3^x = 3^3 \).
03

Equate the Exponents

Since the bases of the exponents are equal, their exponents must also be equal. Therefore, we have: \( x = 3 \).

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Key Concepts

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

Logarithmic Properties
Logarithmic properties are essential tools when dealing with equations involving exponents, especially when trying to simplify complex expressions. Logs can be thought of as the inverse operations of exponentiation, converting multiplication into addition and powers into products.
  • Power Rule: This rule states that \(\log(a^b) = b \, \log(a)\), allowing us to bring down exponents as coefficients, which makes simplifying exponential expressions easier.
  • Change of Base: The change of base formula \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\) can be useful when you want to convert between different bases of logarithms.
In solving exponential equations, logarithmic properties like these come into play as they help manipulate terms into forms that are more workable. For example, by recognizing that \(e^{a\ln n} = n^a\), we can rewrite complex exponential expressions into a simpler form, making solving them straightforward.
Exponential Properties
Understanding exponential properties is crucial for solving equations involving powers and exponents. Exponential functions deal with expressions where the variable is in the exponent position, which often requires particular rules or properties to manipulate and solve.
  • Same Base Rule: If \(a^m = a^n\), then \(m = n\). This suggests that if the bases are identical, the exponents must also be equal for the equation to hold true.
  • Product Property: \(a^m \cdot a^n = a^{m+n}\). This property allows us to add exponents when their bases are the same.
  • Quotient Property: \(\frac{a^m}{a^n} = a^{m-n}\), which simplifies expressions by subtracting exponents.
In the context of the exercise, recognizing that the equation \(3^x = 27\) translates to an exponential equation allows us to proceed by expressing 27 as \(3^3\), thereby matching bases.
Equating Exponents
Equating exponents is a direct approach used in solving exponential equations once the bases on both sides of the equation are the same. This technique implies that if the exponential expressions have identical bases, you can simply equate their exponents.
This property is a logical consequence of the function being one-to-one. In math, a one-to-one function has exactly one output for each unique input, therefore ensuring when \(b^m = b^n\), that \(m = n\).
  • This property is straightforward and works well under the premise that the bases must be the same, allowing for the exponents themselves to solve for the variable.
Applying this to our problem means that after rewriting the expression as \(3^x = 3^3\), we easily find \(x = 3\) by equating the exponents. This simple yet powerful property allows efficient resolution of many exponential equations.

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Most popular questions from this chapter

The loudness of a sound, as experienced by the human ear, is based on its intensity level. A formula used for finding the intensity level \(\alpha\) (in decibels) that corresponds to a sound intensity \(I\) is \(\alpha=10 \log \left(I / I_{0}\right),\) where \(I_{0}\) is a special value of \(I\) agreed to be the weakest sound that can be detected by the ear under certain conditions. Find \(\alpha\) if. (a) \(I\) is 10 times as great as \(I_{0}\) (b) \(I\) is 1000 times as great as \(I_{0}\) (c) \(I\) is \(10,000\) times as great as \(I_{0}\) (This is the intensity level of the average voice.)

Graph \(f\) and \(g\) on the same coordinate plane, and estimate the solution of the equation \(f(x)=g(x)\) $$f(x)=x ; \quad g(x)=-x^{2}-\log _{5} x$$

Exer. \(69-70\) : Bird calls decrease In intensity (loudness) as they travel through the atmosphere. The farther a bird is from an observer, the softer the sound. This decrease in intensity can be used to estimate the distance between an observer and a bird. A formula that can be used to measure this distance is $$ \boldsymbol{I}=\boldsymbol{I}_{\mathbf{0}}-\boldsymbol{2 0} \log \boldsymbol{d}-\boldsymbol{k} \boldsymbol{d} \quad \text { provided } \quad \mathbf{0} \leq \boldsymbol{I} \leq \boldsymbol{I}_{\mathbf{\theta}} $$ where \(I_{0}\) represents the intensity (in decibels) of the bird at a distance of one meter \(\left(I_{0}\) is often known and usually de- \right. pends only on the type of bird), \(I\) is the observed intensity at a distance \(d\) meters from the bird, and \(k\) is a positive constant that depends on the atmospheric conditions such as temperature and humidity. Given \(I_{0 n} I,\) and \(k,\) graphically estimate the distance \(d\) between the bird and the observer. $$I_{0}=60, \quad I=15, \quad k=0.11$$

Sketch the graph of \(f\). $$f(x)=\ln |x|$$

Approximate \(x\) to three significant figures. (a) \(\log x=3.6274\) (b) \(\log x=0.9469\) (c) \(\log x=-1.6253\) (d) \(\ln x=2.3\) (e) \(\ln x=0.05\) (f) \(\ln x=-1.6\)

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