Respuesta :

frika

For the equation [tex]5^{3x+1}=4^{x-5}[/tex] take the [tex]\log[/tex] from both sides:

[tex]\log 5^{3x+1}=\log 4^{x-5},\\ \\(3x+1)\log 5=(x-5)\log 4.[/tex]

Find x:

[tex]3x\log 5+\log 5=x\log 4-5\log 4,\\ \\3x\log 5-x\log 4=-5\log 4-\log 5,\\ \\x(3\log 5-\log 4)=-5\log 4-\log 5,\\ \\x=\dfrac{-5\log 4-\log 5}{3\log 5-\log 4}=\dfrac{5\log 4+\log 5}{\log 4-3\log 5}.[/tex]

Answer: correct choice is D

alinakincsem

Answer:

The correct answer option is [tex]x = \frac{5log4+log5}{log4-3log5}[/tex].

Step-by-step explanation:

We are given the following expression for which we have to make [tex]x[/tex] the subject and solve for it:

[tex]5^{3x+1}= 4^{x-5}[/tex]

Taking log from both sides to get:

[tex]=log 5(3x+1) = log 4 (x-5)[/tex]

Solving the brackets to get:

[tex]3x log5+log 5 = xlog4-5log4[/tex]

[tex]3xlog5+log5 = xlog4 -5log4[/tex]

Taking [tex]x[/tex] as the common to get:

[tex]x(3log5 - log 4) = -5log4 - log5[/tex]

[tex]x = \frac{5log4+log5}{log4-3log5}[/tex]