Question

7B2%7Dx%20%3D%202" id="TexFormula1" title=" log_{2}(3x + 4) - 7 log_{4}{x}^{2} + log_{2}x = 2" alt=" log_{2}(3x + 4) - 7 log_{4}{x}^{2} + log_{2}x = 2" align="absmiddle" class="latex-formula">
simplify​

Answer 1

First of all, we need all logarithms to have the same base. So, we use the formula

$\log_a(b)=\dfrac{\log_c(b)}{\log_c(a)}$

To change the second term as follows:

$\log_4(x^2)=\dfrac{\log_2(x^2)}{\log_2(4)}=\dfrac{\log_2(x^2)}{2}$

Finally, using the property

$\log(a^b)=b\log(a)$

we have

$\dfrac{\log_2(x^2)}{2}=\log_2(x)$

So, the equation becomes

$\log_2(3x+4)-7\log_2(x)+\log_2(x)=2 \iff \log_2(3x+4)-6\log_2(x)=2$

We can now use the formula

$\log(a)-\log(b)=\log\left(\dfrac{a}{b}\right)$

to write the equation as

$\log_2(3x+4)-6\log_2(x)=2 \iff \log_2(3x+4)-\log_2(x^6)=2 \iff \log_2\left(\dfrac{3x+4}{x^6}\right)=2$

Now consider both sides as exponents of 2:

$\dfrac{3x+4}{x^6}=4 \iff 4x^6-3x-4=0$

This equation has no "nice" solution, so I guess the problem is as simplifies as it can be