Question

hw unit 15 - logarithmic and exponential equations: applications
question 32 of 35 (1 point) | question attempt: 1 of unlimited
\ln 5^{k + 1} = \ln 2^k

use the power property of logarithms.

\boxed{k + 1} \ln 5 = k \ln 2

apply the distributive property.

k \ln 5 + \ln 5 = k \ln 2

part: 1 / 3

part 2 of 3

collect k - terms on one side.

k \ln 5 - \square = -\ln 5

(there are some buttons and a box for input)

Explanation:

Step1: Analyze the equation

We start with \( k\ln5 + \ln5 = k\ln2 \). We need to collect the \( k \)-terms on one side. So we subtract \( k\ln2 \) from both sides and also subtract \( \ln5 \) from both sides (or we can think of moving \( k\ln2 \) to the left and \( \ln5 \) to the right). Let's rearrange the equation:

Starting from \( k\ln5 + \ln5 = k\ln2 \)

Subtract \( k\ln2 \) from both sides: \( k\ln5 - k\ln2 + \ln5 = 0 \)

Then subtract \( \ln5 \) from both sides: \( k\ln5 - k\ln2 = -\ln5 \)

Factor out \( k \) on the left: \( k(\ln5 - \ln2) = -\ln5 \), but for the step of collecting \( k \)-terms, we can see that we need to have \( k\ln5 - k\ln2 = -\ln5 \), so the term we subtract from \( k\ln5 \) is \( k\ln2 \). Wait, let's check the given equation in part 2: \( k\ln5 - \square = -\ln5 \)

From \( k\ln5 + \ln5 = k\ln2 \), we can subtract \( k\ln2 \) and \( \ln5 \) from both sides:

\( k\ln5 + \ln5 - k\ln2 - \ln5 = k\ln2 - k\ln2 - \ln5 \)

Simplify left side: \( k\ln5 - k\ln2 \)

Right side: \( -\ln5 \)

But the given form is \( k\ln5 - \square = -\ln5 \), so we can rewrite \( k\ln5 - k\ln2 = -\ln5 \), so the square should be \( k\ln2 \)

Wait, maybe another way: Starting from \( k\ln5 + \ln5 = k\ln2 \)

Subtract \( k\ln2 \) from both sides: \( k\ln5 - k\ln2 + \ln5 = 0 \)

Then subtract \( \ln5 \) from both sides: \( k\ln5 - k\ln2 = -\ln5 \)

But the equation in part 2 is \( k\ln5 - \square = -\ln5 \), so comparing, \( \square = k\ln2 \)

So the missing term is \( k\ln2 \)

Step2: Verify

Let's substitute back. If we have \( k\ln5 - k\ln2 = -\ln5 \), we can factor \( k \) on the left: \( k(\ln5 - \ln2) = -\ln5 \), which is consistent with the previous steps. So the term to fill in the square is \( k\ln2 \)

Answer:

\( k\ln2 \)