Math 3 honors

Not quite!

log₄(2x-1) + log₄(3) = log₄((2x-1)*3)

So we should have

(2x-1)*3 = 5x+12

6x - 3 = 5x + 12

x = 15

Check:

$${{log}}_{{\mathtt{4}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\mathtt{\,\small\textbf+\,}}{{log}}_{{\mathtt{4}}}{\left({\mathtt{3}}\right)} = {\mathtt{3.221\: \!471\: \!747\: \!924\: \!364\: \!5}}$$

$${{log}}_{{\mathtt{4}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{15}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}\right)} = {\mathtt{3.221\: \!471\: \!747\: \!924\: \!364\: \!4}}$$

same log base so rewrite 

2x-1 + 3 = 5x+12 

combine like terms 

2x+2=5x+12 

get x by it self 

3x=10 

solve for x 

x = 10/3