Question

How do i solve for x? *you have to show work*

1.) 2e^2x + 4xe^2x = 0
2.) 2 + 3(4)^(2x-1) = 43

Answer 1

1) Factor the equation first:


`2e^(2x) + 4xe^(2x) = 0`

`e^(2x)(2 + 4x) = 0`

So either
`e^(2x) = 0` or
`2 + 4x = 0`. But
`e^(2x) = 0` can never equal 0! So, we only need to solve for the second equation:


`2 + 4x = 0`

`\bf x = {-(1)/(2)}`

2) First, single out the term with the exponent:


`2 + 3(4^(2x - 1)) = 43`

`3(4^(2x - 1)) = 41`

`4^(2x - 1) = (41)/(3)`

Now, taking the log base 4 of each side gives us


`2x - 1 = \log_4 (41)/(3)`

and we can now solve for x:


`2x - 1 = \log_4 (41)/(3)`

`2x = 1 + \log_4 (41)/(3)`

`2x = \log_4 4 + \log_4 (41)/(3)`

`2x = \log_4 (164)/(3)`

`x = (1)/(2) \log_4 (164)/(3)`

`x = \log_4 \sqrt (164)/(3)`

`\bf x = \log_4 2 \sqrt (41)/(3)`