How do I solve $\log _5 (4x - 6) - 3 = \log _5 (2x -3)$ algebraically and graphically? I been struggling at this question all day. Was told to graph $\log _5 (4x - 6) - 3 =  \log _5 (2x -3)$ algebraically and graphically and list all roots.
Can't seem to find the answer. I thought I got close a couple times, but none of my answers seem to make sense when I actually check it.
 A: Long way first, I didn't notice the multiple,
$$\log_5(4x-6)-3=\log_5(2x-3)$$
$$\log_5(4x-6)-\log_5 125=\log_5(2x-3)$$
$$\log_5\frac{4x-6}{125}=\log_5(2x-3)$$
$$\frac{4x-6}{125}=2x-3$$
$$4x-6=250x-375$$
$$246x=369$$
$$x=\frac{3}{2}$$
But, if you put this in the given equation, you see $\log_50$ which is undefined. So there is no solution.
Or starting from begining, since $4x-6=2(2x-3)$,
$$\log_5(2(2x-3))-3=\log_5(2x-3)$$
$$\log_5(2x-3)+log_52-3=\log_5(2x-3)$$
$$log_5 2=3$$
$$2=5^3=125$$
which is false. There is no solution.
A: $\log_5 (4x-6)-3=\log_5 (2x-3)$
you raise 5 yo the power of both sides of the equality
$5^{\log_5 (4x-6)-3}=5^{\log_5 (2x-3)}$
separate the powers on the left side
$5^{\log_5 (4x-6)}* 5^{-3}=5^{\log_5 (2x-3)}$
the logarithms and exponentials cancel out and you get
$(4x-6)*5^{-3}=2x-3$
from now on it's a just 1st degree equation
$4*5^{-3}x-6*5^{-3}=2x-3$
$(4*5^{-3}-2)x=-3+6*5^{-3}$
so $x=\frac{-3+6*5^{-3}}{(4*5^{-3}-2)}=\frac{3}{2}$
The problem is that in this equation both functions diverge to negative infinity at x=3/2.
If you plot $f(x)=\log_5 (4x-6)-3$ and $g(x)=\log_5 (2x-3)$ into a graphing calculator you will see that both of the functions seem to meet on a vertical line at x=3/2.
Since $\log_5(4x-6)=\log_5(2(2x-3))$
If you look just at the logarithms in both functions you can see that since both have an argument that is multiplied by 2x-3 the arguments will have the same roots, so the logarithms will go to negative infinity on the same point.
