Question

elsas work: log₂x = log₂(3x + 5) + 4 log₂x - log₂(3x + 5) = 4 log₂(3x + 5)x = 4 2⁴ = (3x + 5)x 16 = (3x + 5)x 16(3x + 5) = x 48x + 80 = x 80 = -47x -80/47 = x there is no solution to this equation. elsa solves the logarithmic equation log₂x = log₂(3x + 5) + 4. determine if elsas work is correct or incorrect and justify your decision. for any portion of elsas solution in which there is an error, provide and explain an alternative response. if you wish to include an image in your explanation, you can draw the image on paper, take a picture, and insert it into your response using the image tool.

Explanation:

Step1: Use logarithm property

The correct use of the logarithm property $\log_aM-\log_aN = \log_a\frac{M}{N}$. Starting from $\log_2x=\log_2(3x + 5)+4$, we can rewrite it as $\log_2x-\log_2(3x + 5)=4$, then $\log_2\frac{x}{3x + 5}=4$.

Step2: Convert to exponential form

By the definition of logarithms, if $\log_ab=c$, then $b = a^c$. So from $\log_2\frac{x}{3x + 5}=4$, we get $\frac{x}{3x + 5}=2^4=16$.

Step3: Cross - multiply

Cross - multiplying $\frac{x}{3x + 5}=16$ gives $x=16(3x + 5)$.

Step4: Expand and solve for x

Expand the right - hand side: $x = 48x+80$. Then, move the $x$ terms to one side: $x-48x=80$, so $- 47x=80$, and $x=-\frac{80}{47}$.

Step5: Check the domain

For the original logarithmic equation $\log_2x$ and $\log_2(3x + 5)$, the arguments $x>0$ and $3x + 5>0$. When $x =-\frac{80}{47}$, it does not satisfy $x>0$. So there is no solution.

Answer:

There is no solution to this equation.