Question

Check for extraneous solutions.
2•log₂ (3x-7)=10

Answer 1

`2 log_(2)(3x - 7) = 10`

`log_(2)(3x - 7) = 5`

`3x - 7 = {2}^(5)`

`3x - 7 = 32`

`3x = 39`

`x = 13`

Checking this:

`2 log_(2)(3(13) - 7) = 10`

`2 log_(2)(39 - 7) = 10`

`2 log_(2)(32) = 10`

`2(5) = 10`

32 is the correct solution.

Answer 2

The given logarithmic equation 2•log₂ (3x-7)=10 is solved by first dividing by 2, rewriting in exponential form, and then solving for x, resulting in x=13. The solution is checked and confirmed to not be extraneous by substituting it back into the original equation.

Step-by-step explanation:

To solve the equation 2•log₂ (3x-7)=10, we need to first transform it so that we can isolate the logarithmic term. By dividing both sides of the equation by 2, we get log₂ (3x-7)=5. Next, we can rewrite the equation in exponential form: 2¹⁵ = 3x - 7. This translates to 32 = 3x - 7. Adding 7 to both sides gives us 3x = 32 + 7. Simplifying further, we have 3x = 39.

Now, we divide both sides by 3 to solve for x, resulting in x = 13. Finally, we must check for extraneous solutions by substituting back into the original equation. Since 3(13) - 7 = 32, and taking the logarithm base 2 of 32 indeed gives us 5, we confirm that x = 13 is a valid solution and not extraneous.

It is important to note that whenever we deal with logarithmic equations, checking for extraneous solutions is an essential step, because the operations involved in solving these equations may produce solutions that do not satisfy the original equation.