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

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Answer 1 · 2018-03-09T23:52:07

Real solution: $x = \frac{1}{3} (4 + \frac{ln 7}{ln 4})$ $x = 1/3(4+ln 7/ln 4)$

Complex solutions: $x = \frac{1}{3} (4 + \frac{ln 7}{ln 4} + \frac{2 k π}{ln 4} i)$ $x = 1/3(4+ln 7/ln 4 + (2kpi)/ln 4 i)" "$ for any integer $k$ $k$

Given:

$4^{3 x - 4} = 7$ $4^(3x-4) = 7$

Take natural logs of both sides to get:

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

Divide both sides by $ln 4$ $ln 4$ to get:

$3 x - 4 = \frac{ln 7}{ln 4}$ $3x-4 = ln 7/ ln 4$

Add $4$ $4$ to both sides to get:

$3 x = 4 + \frac{ln 7}{ln 4}$ $3x = 4+ln 7/ln 4$

Divide both sides by $3$ $3$ to get:

$x = \frac{1}{3} (4 + \frac{ln 7}{ln 4})$ $x = 1/3(4+ln 7/ln 4)$

That's the real valued solution.

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

$4^{\frac{2 k π i}{ln 4}} = {(e^{ln 4})}^{\frac{2 k π i}{ln 4}} = e^{2 k π i} = 1$ $4^((2kpii)/ln 4) = (e^(ln 4))^((2kpii)/ln 4) = e^(2kpii) = 1$

for any integer $k$ $k$

So other solutions are given by:

$3 x - 4 = \frac{ln 7 + 2 k π i}{ln 4}$ $3x-4 = (ln 7+2kpii)/ln 4$

Hence:

$x = \frac{1}{3} (4 + \frac{ln 7}{ln 4} + \frac{2 k π}{ln 4} i)$ $x = 1/3(4+ln 7/ln 4 + (2kpi)/ln 4 i)$

Can someone solve this logarithmic equation? 43x−4=74^(3x-4)=7 i just cant find an answer