Problem 12
Question
$$ e^{3 x}=e^{2 x-1} $$
Step-by-Step Solution
Verified Answer
\(x = -1\)
1Step 1: Set the Equation with Equal Bases
Since the bases of the exponents on both sides of the equation are the same (both are base \(e\)), you can set the exponents equal to each other. This results in the equation: \[3x = 2x - 1\].
2Step 2: Solve for \(x\)
You can now solve for \(x\). Start by subtracting \(2x\) from both sides of the equation: \(3x - 2x = -1\). This simplifies to \(x = -1\).
Key Concepts
Equal BasesSolving for VariablesLogarithmic Functions
Equal Bases
When solving exponential equations, having equal bases on both sides of the equation is very beneficial. This scenario allows us to set the exponents equal to each other without worrying about converting or re-arranging the base itself. Let's take a simple example: if we have an equation like \( a^{m} = a^{n} \), where both sides have the same base \(a\), we can deduce that the exponents must be equal too, which gives us \(m = n\). This simplifies the process though it assumes that the base is not zero or one, as those are special cases and could lead to multiple solutions.
- For the equation \(e^{3x}=e^{2x-1}\), both sides have base \(e\).
- Since the bases are the same, the problem reduces to setting the exponents equal: \(3x = 2x - 1\).
Solving for Variables
Once the bases are equal and the exponents are set equal, the next step is to solve for the variable in the equation. In our example, we are trying to find the value of \(x\) in the equation \(3x = 2x - 1\).
- First, eliminate one of the variables by moving it to the other side of the equation. Subtract \(2x\) from both sides: \(3x - 2x = -1\).
- This simplifies to \(x = -1\). It's important to simplify the equations step by step to avoid mistakes.
Logarithmic Functions
In some cases, you might need to solve exponential equations when equal bases aren't immediately apparent or if the bases are different. This is where logarithmic functions come into play. Logarithms are the inverse operations of exponentiation, and they provide a path to isolate variables within exponents.For example, if you have an equation of the form \(a^{x} = b\), you can apply logarithms to both sides to solve for \(x\). This would give you \(x = \log_{a}(b)\), or if you use natural logarithms: \(x = \frac{\ln(b)}{\ln(a)}\).
- Logarithms convert multiplicative relationships into additive ones, making complex equations easier.
- They help in solving for variables when dealing with different bases or when simplification isn't straightforward.
Other exercises in this chapter
Problem 11
Sketch the graph of \(f\) if \(a=2\). (a) \(f(x)=a^{x}\) (b) \(f(x)=-a^{x}\) (c) \(f(x)=3 a^{x}\) (d) \(f(x)=a^{x+3}\) (e) \(f(x)=a^{x}+3\) (f) \(f(x)=a^{x-3}\)
View solution Problem 12
Write the expression as one logarithm. $$ 5 \log _{a} x-\frac{1}{2} \log _{a}(3 x-4)-3 \log _{a}(5 x+1) $$
View solution Problem 12
Exer. 11-24: Find the exact solution, using common logarithms, and a two- decimal-place approximation of each solution, when appropriate. $$ 4^{2 x+3}=5^{x-2} $
View solution Problem 12
Exer. 11-12: Change to logarithmic form. (a) \(10^{4}=10,000\) (b) \(10^{-2}=0.01\) (c) \(10^{x}=38 z\) (d) \(e^{4}=D\) (e) \(e^{0.1 t}=x+2\)
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