Question

Explain why you could use a base 2 or a base 4 to solve the problem above and get the same answer. Show your work and explain.

Answer 1

Step 1. The expression that we have is:

`4^x=16^(x-2)`

To prove that we get the same answer if we use a logarithm base 2 and a logarithm base 4, we will solve the equation using both and check that the result is the same.

Step 2. To solve using a base 2 logarithm, we apply it to both sides of the equation:

`log_2(4^x)=log_2(16^(x-2))`

Using the following property of logarithms:

`log(x^m)=mlog(x)`

The expression is simplified as follows:

`xlog_2(4)=(x-2)log_2(16)`

The base 2 logarithm of 4 and 16 is:

`\begin{gathered} log_2(4)=2 \\ log_2(16)=4 \end{gathered}`

Substituting these values into the equation:

`\begin{gathered} x(2)=(x-2)(4) \\ Simplifying: \\ 2x=4x-8 \end{gathered}`

Solving for x:

`\begin{gathered} 2x-4x=-8 \\ -2x=-8 \\ x=(-8)/(-2) \\ \boxed{x=4} \end{gathered}`

Step 3. Now we repeat the process but this time we use the logarithm base 4.

The expression is:

`4^(x)=16^(x-2)`

Applying logarithm base 4 to both sides:

`log_4(4^x)=log_4(16^(x-2))`

Simplifying:

`xlog_4(4)=(x-2)log_4(16)`

The base 4 logarithm of 4 and 16 is:

`\begin{gathered} log_4(4)=1 \\ log_4(16)=2 \end{gathered}`

Substituting these values into our equation:

`\begin{gathered} x(1)=(x-2)(2) \\ Simplifying\text{ and solving for x:} \\ x=2x-4 \\ x-2x=-4 \\ -x=-4 \\ \boxed{x=4} \end{gathered}`

Answer: We have proven that we get the same result using a base 2 logarithm and a base 4 logarithm.