I am to solve $\frac{3}{log_2(10)}-log(x-9)=log(44)$.
The textbook solution provided is $x=\frac{101}{11}$ but I am unable to arrive at a solution.
Here's as far as I was able to get:
$$\frac{3}{log_2(10)}-log(x-9)=log(44)$$
I want to convert the denominator $log_2$ into the same base (10) as the others:
$$log_2(10)=\frac{log(10)}{log(2)}$$
Now swapping this in I have:
$$\frac{3}{(\frac{log(10)}{log(2)})}-log(x-9)=log(44)$$
Which I think I can rewrite as:
$$\frac{3}{log(\frac{10}{2})}-log(x-9)=log(44)$$
Which can be simplified to:
$$\frac{3}{log(5)}-log(x-9)=log(44)$$
This is as far as I got. I don't know where to go next or if I'm on the right track. The 3 in the numerator is what's stumping me, I don't know how to handle it.
How can I solve this equation to arrive at $x=\frac{101}{11}$?
More granular baby steps much appreciated.
[EDIT]
From a commenter, I'm working through an open online textbook, exercise number 63 at the bottom of this page. The corresponding solution for chapter 66, exercise 63 is on this page.
Best Answer
Remember that in your exercise $\log$ means $\log_{10}$. So you have $\log(10)=1$.
Now,
\begin{align} \frac{3}{\log_2(10)}-\log(x-9)&=\log(44)\\ \log(x-9)&=\frac{3}{\log_2(10)}-\log(44)&\textrm{(put the unknown to one side)}\\ \log(x-9)&=\frac{3\log(2)}{\log(10)}-\log(44)&(\log_2(10)=\frac{\log(10)}{\log(2)})\\ \log(x-9)&=3\log(2)-\log(44)\\ \log(x-9)&=\log\frac{2^3}{44}\\ x-9&=\frac{8}{44}\\ x&=\frac{101}{11} \end{align}
Notes.
In general, $$ \log \frac{b}{a}\ne \frac{\log b}{\log a} $$