State domain, range, vertical asymptote and $x,y$ intercepts for $\log_4(x-1)+1$

Question

State domain, range, vertical asymptote and $x,y$ intercepts for $\log_4(x-1)+1$

I have the solution in the answers section of my book:

Domain: $(1,\infty)$; Range: $(−\infty,\infty)$; Vertical asymptote: $x=1$;

$x$-intercept: $(5/4,0)$; $y$-intercept: Does not exist

I am able to figure out everything except the $x$-intercept.

Domain:

$$x-1 > 0$$
$$x>1$$

Domain therefore is $(1,\infty)$ and the vertical asymptote is 1.
Range is $(-\infty, \infty)$ and there is no $y$ intercept because it's a log function.

For the parent function $\log_4(x)$ the x-intercept is $(1,0)$. How can I arrive at $(\frac{5}{4},0)$?

Granular, baby steps appreciated.

Answer 1

Looks all fine. You can solve the equation straightforward.

$\log_4(x-1)+1=0$

$\log_4(x-1)=-1$

Writing both sides into the exponent with base $4$.

$4^{\log_4(x-1)}=4^{-1}$

$x-1=\frac14$

$\vdots$