Question

1. the equation $2^{x + 3} = 3^{2x - 1}$ has the solution:

a. $x = \frac{-\log 3 - 3\log 2}{\log 2 - 2\log 3}$
b. $x = \frac{2}{3}$
c. $x = 1$
d. no solution

Explanation:

Step1: Take log on both sides

Take the logarithm (we can use common logarithm or natural logarithm, here we use common logarithm for simplicity) of both sides of the equation \(2^{x + 3}=3^{2x-1}\). By the property of logarithms \(\log a^{b}=b\log a\), we get:
\(\log(2^{x + 3})=\log(3^{2x-1})\)
\((x + 3)\log2=(2x - 1)\log3\)

Step2: Expand both sides

Expand the left - hand side and the right - hand side of the equation:
\(x\log2+3\log2 = 2x\log3-\log3\)

Step3: Collect like terms

Move the terms with \(x\) to one side and the constant terms to the other side. Subtract \(2x\log3\) from both sides and subtract \(3\log2\) from both sides:
\(x\log2-2x\log3=-\log3 - 3\log2\)

Step4: Factor out \(x\)

Factor out \(x\) from the left - hand side of the equation:
\(x(\log2 - 2\log3)=-\log3-3\log2\)

Step5: Solve for \(x\)

Divide both sides of the equation by \((\log2 - 2\log3)\) to solve for \(x\):
\(x=\frac{-\log3 - 3\log2}{\log2-2\log3}\)

Answer:

A. \(x = \frac{-\log3 - 3\log2}{\log2-2\log3}\)