How to solve for log? I have this equation which is just a part of a bigger equation, but I would like to simplify it:
$\log(x)= 2+\log(b)$, where $b=3$.
How can I solve for $x$?.
Will anyone help me?
 A: $log(x)=2+log(3)$, $x=\exp\left(2+\log(3)\right)=3e^2$.
Obviously, this is true, if your function $\log(y)$ is the natural logarithm.
A: It's 
$$
log(x) = 2 + log(b) \Rightarrow x = \alpha^{2+log(b)} = \alpha^{2}\alpha^{log(b)} = 
\alpha^{2}b
$$
where $\alpha$ is the base of the logarithm. If $a=e$ and $b=3$, then $log(x)=3e^n$.
A: If a is the base of the log then
log(x)=2+log(b)
      = log(a^2) + log(b)
      = log(b*a^2)
so x= b*a^2
if b= 3 and a is e ( natural log) then x=3*e^2
if a = 10 (log tables, etc) then x=300
A: There are a few ways to do this one. What you do first is get the logs on one side. Subtracting $\log(b)$ from both sides gives us:
$$ \log(x) - \log(b) = 2.$$
Since we said $b=3$, we have:
$$\log(x) - \log(3) = 2.$$
I'm not sure how familiar you are with properties of logs, but we have a "quotient" property for logs that says
$$ \log(A) - \log(B) = \log\left( \dfrac{A}{B} \right).$$
If we use this with our equation, we get:
$$\log\left( \dfrac{x}{3} \right) = 2.$$
Now, we have to get rid of the log. I'm going to assume log means "log base $10$" To do this, we use an inverse property for logs that says:
$$10^{log(x)} = x.$$
Take a base of $10$ on both sides:
$$10^{\log\left( \dfrac{x}{3} \right)} = 10^2.$$
Using the inverse property, the log and the $10$ cancel out, and we are left with
$$\dfrac{x}{3} = 10^2.$$
Finally, solve for x to get...
