The equation $22\log(92x+40.66)=38.9$
steps so far
$\log(92x+40.66)=\frac{38.9}{22}$
to eliminate log, do I have to apply the opposite of log? Not sure what that is.
The equation $22\log(92x+40.66)=38.9$
steps so far
$\log(92x+40.66)=\frac{38.9}{22}$
to eliminate log, do I have to apply the opposite of log? Not sure what that is.
$22\log(92x+40.66)=38.9\Longrightarrow \log(92x+40.66)=\frac{38.9}{22}\Longrightarrow$
$92x+40.66=e^{\frac{38.9}{2}}\Longrightarrow \,\,\ldots$
If by $\,\log\,$ you mean logarithm in base $\,10\,$ just change $\,e\,$ for $\,10\,$
In general, $\log_{b}u=v \iff b^v=u.$
Apply this fact. :-)
You might have heard that $ b^{\log_b(x)} = x. $ Most people that I am aware of just write $\log$ when they mean $\log_{10}$. So that would mean that $ 10^{\log(x)} = x. $ Now you have the equation $ \log(92x+40.66)=\frac{38.9}{22} $ and to get rid of the $\log$ you can $ 10^{\log(92x+40.66)}=10^{\frac{38.9}{22}}. $ This is equivalent to $ 92x + 40.66 = 10^{\frac{38.9}{22}} $ and this is just a linear equation that you probably know how to solve (on both sides: subtract $40.66$ and then divide by $92$).