Question

Solve for \(x\) accurate to three decimal places. (a) \(e^{\ln 2 x}=12\) (b) \(\ln e^{-x}=0\)

Short answer

The solution to part (a) of the exercise is \(x = 6\) and the solution to part (b) is \(x = 0\)

Step-by-step solution

Part (a): Step 1 - Apply the logarithmic and exponential identity

The first step in solving the equation \(e^{ln 2x}=12\) is to use the logarithmic and exponential identity \(e^{ln a} = a\). Here, \(a\) represents \(2x\), so \(e^{ln 2x}\) simplifies to \(2x\). The equation can then be rewritten as \(2x = 12\).

Part (a): Step 2 - Solve for \(x\)

After the equation is simplified to \(2x = 12\), we can solve for \(x\) by dividing both sides of the equation by 2. Therefore, \(x = 12 / 2 = 6\).

Part (b): Step 1 - Apply the logarithmic and exponential identity

Similar to part (a), the equation \(\ln e^{-x} = 0\) is simplified using the logarithmic and exponential identity \(ln(e^a) = a\). Here, \(a\) represents \(-x\), so \(\ln e^{-x}\) simplifies to \(-x\). The equation can therefore be rewritten as \(-x=0\).

Part (b): Step 2 - Solve for \(x\)

After the equation is simplified to \(-x = 0\), we can easily see that the only value for \(x\) that satisfies this equation is \(0\), so \(x=0\).