Hello, petebob6!
Can't tell what you did, but sorry, your answer is wrong . . .
Solve for \(\displaystyle x:\;\;\log(3x\,-\,2)\,-\,\log(2x\,-\,3)\:=\:1\)
We have:
.\(\displaystyle \L\log\left(\frac{3x\,-\,2}{2x\,-\,3}\right)\:=\:1\:=\:\log(10)\)
Take anti-logs:
.\(\displaystyle \L\frac{3x\,-\,2}{2x\,-\,3}\:=\:10\;\;\Rightarrow\;\;3x\,-\,2\:=\:10(2x\,-\,3)\)
. . \(\displaystyle \L3x\,-\,2\:=\:20x\,-\,30\;\;\Rightarrow\;\;17x\,=\,28\;\;\Rightarrow\;\; x\,=\,\frac{28}{17}\)
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Check
\(\displaystyle \L\log\left(3\cdot\frac{28}{17}\,-\,2\right)\,-\,\log\left(2\cdot\frac{28}{17}\,-\,3\right)\:=\:\log\left(\frac{50}{17}\right)\,-\,\lig\left(\frac{5}{17}\right)\)
. . . \(\displaystyle \L=\:\log\left(\frac{\frac{50}{17}}{\frac{5}{17}}\right)\:=\:\log(10)\:=\:1\)
. . . . . Yay!