Hello there. To solve this question, we'll have to remember some properties about exponential equations.
Given the following equation:
[tex]2^{3x-4}=5\cdot3^{-x+4}[/tex]To solve it, take the base 10 logarithm on both sides of the equation:
[tex]\log(2^{3x-4})=\log(5\cdot3^{-x+4})[/tex]Apply the following properties:
[tex]\begin{gathered} 1.\text{ }\log(a\cdot\,b)=\log(a)+\log(b) \\ \\ 2.\text{ }\log(a^b)=b\cdot\log(a) \end{gathered}[/tex]Therefore we get
[tex](3x-4)\cdot\log(2)=\log(5)+(4-x)\cdot\log(3)[/tex]Apply the FOIL, such that we get
[tex]3\log(2)\,x-4\log(2)=\log(5)+4\log(3)-\log(3)\,x[/tex]Add log(3) x + 4 log(2) on both sides of the equation, such that we get
[tex]3\log(2)\,x+\log(3)\,x=\log(5)+4\log(3)+4\log(2)[/tex]Rewrite
[tex]\begin{gathered} 3\log(2)=\log(2^3)=\log(8) \\ \\ 4\log(3)=\log(3^4)=\log(81) \\ \\ 4\log(2)=\log(2^4)=\log(16) \\ \end{gathered}[/tex]Therefore we get
[tex]\begin{gathered} (\log(8)+\log(3))x=\log(5)+\log(81)+\log(16) \\ \\ \Rightarrow\log(24)\,x=\log(6480) \end{gathered}[/tex]Divide both sides of the equation by a factor of log(24)
[tex]x=\dfrac{\log(6480)}{\log(24)}[/tex]This is the answer to this question and it is approximately equal to:
[tex]x\approx2.7615\cdots[/tex]