Question

Can someone solve this logarithmic equation? `4^(3x-4)=7` i just cant find an answer

Answer 1

Real solution: `x = 1/3(4+ln 7/ln 4)`

Complex solutions: `x = 1/3(4+ln 7/ln 4 + (2kpi)/ln 4 i)" "` for any integer `k`

Explanation:

Given:

`4^(3x-4) = 7`

Take natural logs of both sides to get:

`(3x-4) ln 4 = ln 7`

Divide both sides by `ln 4` to get:

`3x-4 = ln 7/ ln 4`

Add `4` to both sides to get:

`3x = 4+ln 7/ln 4`

Divide both sides by `3` to get:

`x = 1/3(4+ln 7/ln 4)`

That's the real valued solution.

If we are interested in the complex solutions, then note that:

`4^((2kpii)/ln 4) = (e^(ln 4))^((2kpii)/ln 4) = e^(2kpii) = 1`

for any integer `k`

So other solutions are given by:

`3x-4 = (ln 7+2kpii)/ln 4`

Hence:

`x = 1/3(4+ln 7/ln 4 + (2kpi)/ln 4 i)`