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 $92$).