Hello aligator0207 Originally Posted by
aligator0207 What happened to the (-1) in problem 1? And at the end, do I put that into my calculator?
I suspect that you should have put brackets around the 2x-1 and that the original equation was actually:$\displaystyle 3^{2x-1}=5$
and, if so, the working should be:$\displaystyle (2x-1)\log3=\log5$
$\displaystyle \Rightarrow 2x-1 = \frac{\log5}{\log3}$
$\displaystyle \Rightarrow x = \frac12\left(1+\frac{\log5}{\log3}\right)$$\displaystyle =\frac{\log3+\log5}{2\log3}$
$\displaystyle =\frac{\log15}{\log9}$
$\displaystyle =1.2325$
Grandad