Equation with Logarithm: $\log_x3+\log_x12 = 2$ Given is the equation:
$$\log_x3+\log_x12 = 2$$
How do I solve it? My idea was to use the formula $\log_a(b) = \frac{\ln b}{\ln a}$ but that does not seem to help here.
 A: Hint: $\log_x3 + \log_x12 = \log_x (3 \times 12) = \log_x 36$.
A: Your method is feasible indeed!
Using the formula $\log_ab=\frac{\ln a}{\ln b}$, we get $$\frac{\ln 3}{\ln x}+\frac{\ln 12}{\ln x}=2 \\ \frac{\ln 3+\ln 12}{\ln x}=2 \\ \frac{\ln(3\times 12)}{\ln x}=2 \\ \ln x=\frac 12\times \ln36 \\ \ln x=\ln{36^\frac 12} \\ \ln x = \ln 6 \\ x=6$$
Of course, change of base to natural logarithm is not a must in this question as in the answer given by @Ross B. But I would like to tell you that your approach is definitely fine.
A: Maybe this way is easier to understand:
$$\log_x3+\log_x12=2$$
$$\therefore\quad x^{\log_x3+\log_x12}=x^2$$
$$\therefore\quad x^{\log_x3}\cdot x^{\log_x12}=x^2$$
$$\therefore\quad 3\cdot12=x^2$$
$$\therefore\quad x^2=36=6^2$$
Since $x$ must be positive (since $\log_x$ must be defined), you get $x=6$.
A: my answer, 
$\log_x3+\log_x12=2$
$\log_x(3\times 12)=2\log_xx$
$\log_x(36)=\log_x(x^2)$
compare the numbers on both the sides, 
$36=x^2\ \ \ \ $ forall $(0<x<1$ or $x>1)$ 
$x=6$
