\begin{equation*} \log_ax = \lvert x+1 \rvert + \lvert x-5 \rvert. \end{equation*} I don't even know how to approach this one, any hints would be amazing.
I tried separating into two cases, where $0<x<5$ and $x>5$. The first pretty much limits a to every positive number as far as I have seen, the latter I don't know how to solve.
Answer is supposed to be
$(0,1)$ and $5^{\frac 16}$
On
For $0< x <5$, $$ \log_ax = x+1-x+5=6 $$ So, $$ a=\sqrt[6]{x} $$
Therefore, $$ 0<a<5^\dfrac{1}{6}\tag{1} $$
For $x>5$, $$ \log_ax = x+1+x-5=2x-4 $$ $$ \log_ax=2x-4 $$ $$ a=\sqrt[2x-4]{x} $$ Using graph, $$ 1<a<5^\dfrac{1}{6}\tag{2} $$
At $x=5$, $$ \log_a5 = 1+5=6 $$
Therefore, $$ a=5^{\frac 16} $$ From equation 1 and 2, $a=5^{\frac 16}$ and $0<a<1$.
On
The graph of the function has the shape of a truncated V, being constantly equal to 6 on the interval [-1, 5] and greater than 6 elsewhere. This is a convex function and so is (in the opposite direction) the logarithm of base a (positive and distinct of 1). Hence in order to have a unique solution, $log_ax$ must touch tangentially f at the point (5, 6) or cuts f to a single point. Then, by convexity, the possible values of a are positive and less than 1.
The point of tangency is given by $log_a5 = 6$ hence $a=\sqrt[6]5$ = 1.307660 and for greater values of a there are no common points. On the other hand if a > 1 and $log_ax$ cuts f, there are two points (by convexity and because the line to the right side increase more than the log). The other possible values, 0 < a < 1, have the positive axis OY as asymptote given then a unique point of ordinate 6.
Hint: Plot the expression $|x+1|+|x-5|$. You will see that for $x \leq -1$, it is a line with slope $-2$; for $-1 \leq x \leq 5$, it is a horizontal line at $y = 6$; for $x \geq 5$, it is a line with slope $2$. It looks a bit like a playground swing.
For any given $a$, the locations at which $\log_a x = |x+1|+|x-5|$ are the places where the graph of $\log_a x$ intersects the "playground swing" plot. Now consider that $\log_a x$ is an increasing function of $x$ for any $a$. For most such $a$, those two plots will intersect in either zero points, or two points. But for one value of $a$, the increasing $\log_a x$ function will intersect the "playground swing" in only one point. What point $(x^*, y^*)$ would that be? Figure that out, and then you can solve for $a$ using
$$ a^{y^*} = x^* $$
which yields
$$ a = (x^*)^{(1/y^*)} $$