Respuesta :
First of all, we need all logarithms to have the same base. So, we use the formula
[tex]\log_a(b)=\dfrac{\log_c(b)}{\log_c(a)}[/tex]
To change the second term as follows:
[tex]\log_4(x^2)=\dfrac{\log_2(x^2)}{\log_2(4)}=\dfrac{\log_2(x^2)}{2}[/tex]
Finally, using the property
[tex]\log(a^b)=b\log(a)[/tex]
we have
[tex]\dfrac{\log_2(x^2)}{2}=\log_2(x)[/tex]
So, the equation becomes
[tex]\log_2(3x+4)-7\log_2(x)+\log_2(x)=2 \iff \log_2(3x+4)-6\log_2(x)=2[/tex]
We can now use the formula
[tex]\log(a)-\log(b)=\log\left(\dfrac{a}{b}\right)[/tex]
to write the equation as
[tex]\log_2(3x+4)-6\log_2(x)=2 \iff \log_2(3x+4)-\log_2(x^6)=2 \iff \log_2\left(\dfrac{3x+4}{x^6}\right)=2[/tex]
Now consider both sides as exponents of 2:
[tex]\dfrac{3x+4}{x^6}=4 \iff 4x^6-3x-4=0[/tex]
This equation has no "nice" solution, so I guess the problem is as simplifies as it can be
