Originally Posted by
HallsofIvy Of course he has. He has x equal to a specific number, $\displaystyle ln_3(2)+ 1$. Perhaps you don't like the way the number is written, but that has nothing to do with "solving". Are you saying that he should solve for $\displaystyle \frac{log(2)}{log(3)}+ 1= \frac{0.3010}{0.47712}+ 1= 0.63093+ 1= 1.06093$? Because that is NOT the correct answer- it is only approximately correct.
It is the same as saying that the solutions to $\displaystyle x^2= 2$ are $\displaystyle \sqrt{2}$ and $\displaystyle -\sqrt{2}$, NOT "1.41421" and "-1.41421".