<s>[INST] Solve $\log_{4}3x+1,+\log_{3}16x+1=\log_{2}12x+4$ Here is a basic logarithmic problem that I am struggling to solve:
$\log_{4}(3x+1)+\log_{3}(16x+1)=\log_{2}(12x+4)$
Here is my step:
$\log_{4}(3x+1)+\log_{4}(\frac{16x+1}{3})=\log_{2}(12x+4)$
Since the bases are the same now, we can multiply the arguments.
$\log_{4}(3x+1)*(\frac{16x+1}{3})=\log_{2}(12x+4)$
$\log_{4}\frac{48x^2+19x+1}{3}=\log_{2}(12x+4)$
The next thing to do is to change the base of 2 to base 4 on the right side.
Therefore,
$\log_{4}\frac{48x^2+19x+1}{3}=\log_{4}\frac{12x+4}{\frac{1}{2}}$
The right side can be simplified further, so
$\log_{4}\frac{48x^2+19x+1}{3}=\log_{4}24x+8$
Since the bases are the same on both sides, the arguments are also the same.
$\frac{48x^2+19x+1}{3}=24x+8$
$48x^2+19x+1=72x+24$
$48x^2-53x-23=0$
Using the quadratic formula, I get
$x=\frac{-(-53)\pm\sqrt{(-53)^2-4*48*-23}}{2*48}$
$x=\frac{53\pm\sqrt{2809+4116}}{96}$
$x=\frac{53\pm\sqrt{7225}}{96}$
$x=\frac{53\pm85}{96}$
$x=\frac{23}{16}$
or
$x=\frac{-1}{3}$
Since log argument cannot be negative, the second solution will not work. But when I checked the solution, $x=5$, which is not the answer that I have. A seemingly easy problem is driving me crazy. Can anyone point out what I'm doing wrong?
 [/INST] The issue with your solution, as already noted in the comments, is when you converted $$\log_3(16x+1)$$ to $$\log_4\left(\frac{16x+1}3\right)$$
The log base rule states that:
$$\log_ba=\frac{\log_da}{\log_db}$$
In this case, $$\log_3(16x+1)$$
would convert to $$\frac{\log_4(16x+1)}{\log_43}$$
which is not equal to $$\log_4\left(\frac{16x+1}3\right)$$
Take, for example, $$\frac{\log_4{16}}{\log_44}=\frac21=2$$
But,
$$\log_4\left(\frac{16}4\right)=\log_44=1$$
Which is obviously not the same as $2$.
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