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Chapter 3: Problem 35

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$2^{x}=3^{x+1}$$

Short Answer

Expert verified
The approximate solution to the equation \(2^x = 3^{x+1}\) is \(x \approx -1.631\).

Step by step solution

01

Take the natural log of both sides

Taking the natural log of both sides of the equation gives us \(\ln(2^x) = \ln(3^{x+1})\). By using the property of logarithms \( \ln(a^b) = b \ln(a)\), we can bring the exponents down: \(x \ln(2) = (x+1) \ln(3)\).
02

Distribute and rearrange terms

From here we can distribute the logarithm on the right-hand side: \(x \ln(2) = x \ln(3) + \ln(3)\). After this, we want to get all terms with \(x\) on one side of the equation and the constants on the other side which results in: \(x \ln(2) - x \ln(3) = \ln(3)\). Then, we can factor out \(x\) to simplify further: \(x(\ln(2) - \ln(3)) = \ln(3)\).
03

Solve for x

Lastly, we just need to isolate \(x\), so we can divide both sides of the equation by \((\ln(2) - \ln(3))\) to solve for \(x\): \(x = \frac{\ln(3)}{\ln(2) - \ln(3)}\). The solution can then be approximated using a calculator.

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