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Solve each equation. Approximate solutions to three decimal places. $$ 4^{x-2}=5^{3 x+2} $$

Short Answer

Expert verified
x \approx -1.742

Step by step solution

01

- Take the natural logarithm on both sides

Apply the natural logarithm \(\text{ln}\text{ln}\)) to both sides of the equation: \[ \ln(4^{x-2}) = \ln(5^{3x+2}) \]
02

- Apply the power rule for logarithms

Use the power rule \(\text{ln}(a^b) = b \cdot \ln(a)\)) to bring down the exponents: \[ (x-2) \ln(4) = (3x+2) \ln(5) \]
03

- Distribute the logarithms

Expand the equation: \[ x \ln(4) - 2 \ln(4) = 3x \ln(5) + 2 \ln(5) \]
04

- Combine like terms

Isolate the terms with \(x\): \[ x \ln(4) - 3x \ln(5) = 2 \ln(4) + 2 \ln(5) \]
05

- Factor out x

Factor out \(x\) on the left-hand side: \[ x (\ln(4) - 3 \ln(5)) = 2 \ln(4) + 2 \ln(5) \]
06

- Solve for x

Divide both sides by \(\ln(4) - 3 \ln(5)\): \[ x = \frac{2 \ln(4) + 2 \ln(5)}{\ln(4) - 3 \ln(5)} \] Compute the approximate solution: \[ x \approx \frac{2(1.386) + 2(1.609)}{1.386 - 3(1.609)} \approx \frac{5.99}{-3.44} \approx -1.742 \]

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Key Concepts

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

Natural Logarithms
Natural logarithms (\text{ln}) are logarithms with a base of the mathematical constant \(e\), which is approximately 2.718. Natural logarithms are commonly used in solving exponential equations due to their properties that make it easier to handle complex expressions.
To solve an exponential equation like \(4^{x-2}=5^{3x+2}\), taking the natural logarithm on both sides helps to linearize the equation, allowing us to work with exponents in a more straightforward manner.
Using natural logarithms simplifies many steps in solving these problems by transforming multiplication into addition and handling large exponents efficiently.
Power Rule for Logarithms
The power rule for logarithms states that for any positive number \(a\) and any real number \(b\), the logarithm of \(a^b\) is equal to \(b\text{ln}(a)\). Mathematically, this is expressed as \(\text{ln}(a^b)=b\cdot\text{ln}(a)\).
In our problem, applying the power rule allows us to bring down the exponents so we can work with linear equations rather than exponential ones.
For example:
  • Starting with \(\text{ln}(4^{x-2})\), applying the power rule transforms it to \((x-2)\cdot\text{ln}(4)\).
  • Similarly, \(\text{ln}(5^{3x+2})\) becomes \((3x+2)\cdot\text{ln}(5)\).
Once the exponents are brought down, solving the equation becomes much simpler.
Isolating Variables
Isolating variables is a critical step in solving algebraic equations. It involves moving terms with the variable of interest to one side of the equation and constants to the other side.
In our scenario, after applying the power rule, we get a linear equation: \((x-2)\text{ln}(4) = (3x+2)\text{ln}(5)\).
We first distribute the logarithms to get: \(x\text{ln}(4) - 2\text{ln}(4) = 3x\text{ln}(5) + 2\text{ln}(5)\).
Then, we combine like terms to isolate \(x\): \(x\text{ln}(4) - 3x\text{ln}(5) = 2\text{ln}(4) + 2\text{ln}(5)\).
Finally, factoring out \(x\) on the left side gives us: \(x (\text{ln}(4) - 3\text{ln}(5)) = 2\text{ln}(4) + 2\text{ln}(5)\).
Approximations to Three Decimal Places
When solving equations, the exact solution might not always be practical or possible to obtain. In such cases, we approximate the solution to a certain number of decimal places.
For our problem, the final step requires calculating an exact numerical value of \(x\). For that, we approximate the natural logarithms to three decimal places:
  • \(\text{ln}(4)\) is approximately 1.386.
  • \(\text{ln}(5)\) is approximately 1.609.
Substituting these values into the equation \( x = \frac{2\text{ln}(4) + 2\text{ln}(5)}{\text{ln}(4) - 3\text{ln}(5)} \), we get:
\( x \frac{2(1.386) + 2(1.609)}{1.386 - 3(1.609)} \approx \frac{5.99}{-3.44} \approx -1.742.\) This gives our approximate solution: \(x \approx -1.742.\). Approximations let us understand the scale of the answer while keeping calculations manageable.

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Most popular questions from this chapter

Inverse functions can be used to send and receive coded information. A simple example might use the function \(f(x)=2 x+5 .\) (Note that it is one-to-one.) Suppose that each letter of the alphabet is assigned a numerical value according to its position, as follows. $$\begin{array}{llllllllll}\mathbf{A} & 1 & \mathbf{G} & 7 & \mathbf{L} & 12 & \mathbf{Q} & 17 & \mathbf{V} & 22 \\\\\mathbf{B} & 2 & \mathbf{H} & 8 & \mathbf{M} & 13 & \mathbf{R} & 18 & \mathbf{W} & 23 \\\\\mathbf{C} & 3 & \mathbf{I} & 9 & \mathbf{N} & 14 & \mathbf{S} & 19 & \mathbf{X} & 24 \\\\\mathbf{D} & 4 & \mathbf{J} & 10 & \mathbf{O} & 15 & \mathbf{T} & 20 & \mathbf{Y} & 25 \\\\\mathbf{E} & 5 & \mathbf{K} & 11 & \mathbf{P} & 16 & \mathbf{U} & 21 & \mathbf{Z} & 26 \\\\\mathbf{F} & 6 & & & & & & & &\end{array}$$ Using the function, the word ALGEBRA would be encoded as $$\begin{array}{lllllll}7 & 29 & 19 & 15 & 9 & 41 & 7\end{array}$$ because \(f(\mathrm{~A})=f(1)=2(1)+5=7, \quad f(\mathrm{~L})=f(12)=2(12)+5=29, \quad\) and so on The message would then be decoded using the inverse of \(f,\) which is \(f^{-1}(x)=\frac{x-5}{2}\). $$f^{-1}(7)=\frac{7-5}{2}=1=\mathrm{A}, \quad f^{-1}(29)=\frac{29-5}{2}=12=\mathrm{L}, \quad \text { and so on }$$ Use \(f(x)=x^{3}+4\) to encode your name, using the above letter/number assignment

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