Evaluate log base 3 of 18, divided by the quantity 2 plus log base 3 of 2.

Step-by-step solution

1. 1

   Read the notation. The original was written with a colon, $\log_3 18:(2+\log_3 2)$, and in this convention "$a:b$" means $a\div b$. So the task is to evaluate the single fraction $\dfrac{\log_3 18}{2+\log_3 2}$.

2. 2

   Factor 18 so that a power of the base appears. The base is $3$, so look for factors of $3$:

   $18=9\cdot 2=3^2\cdot 2$

   This is the whole idea — the denominator already contains a bare $2$ and a $\log_3 2$, which hints that the numerator hides the same two pieces.

3. 3

   Split the logarithm with the product rule. Using $\log_b(MN)=\log_b M+\log_b N$,

   $\log_3 18=\log_3(3^2)+\log_3 2=2+\log_3 2$

   since $\log_3(3^2)=2$ exactly.

4. 4

   Substitute and cancel. The numerator is now identical to the denominator:

   $\frac{\log_3 18}{2+\log_3 2}=\frac{2+\log_3 2}{2+\log_3 2}=1$

   The cancellation is valid because $\log_3 2\approx 0.6309>0$, so the denominator is about $2.63$ and certainly not zero.

5. 5

   Confirm numerically. $\log_3 18=\frac{\ln 18}{\ln 3}=2.6309298$ and $2+\log_3 2=2+0.6309298=2.6309298$. The two agree to seven decimals, so the ratio is exactly $1$.