How to work out value of $a$ and $b$ in exponents? I am a student and I need a step by step solution in working let the value of $a$ and $b$ in this question.

$$1800 = 2^3 × 3^a × 5^b$$

I don't know what steps to take in order to work out what the powers are. 
Thank you and help is appreciated.

Sweep through the prime numbers $2, 3, 5, 7, 9, 13, \dots $ as indicated by @projectilemotion.

First prime: $p=\boxed{2}$
$$
 \frac{1800}{2^{3}} = 255
$$

Decomposition is now $1800 = 2^{3} \times ?$

Next prime: $p=\boxed{3}$
$$
 \underbrace{\frac{225}{3} = 75}_{1} 
\qquad \Rightarrow \qquad
 \underbrace{\frac{75}{3} = 25}_{2}
\qquad \Rightarrow \qquad
 \frac{25}{3} \notin \mathbb{Z}
$$
The prime was used $\color{blue}{2}$ times.

Decomposition is now $1800 = 2^{3} 3^{\color{blue}{2}} \times ?$

Next prime: $p=\boxed{5}$
$$
 \underbrace{\frac{25}{5} = 5}_{1} 
\qquad \Rightarrow \qquad
 \underbrace{\frac{5}{5} = 1}_{2}
\qquad \Rightarrow \qquad
 \frac{1}{5} \notin \mathbb{Z}
$$
The prime was used $\color{red}{2}$ times.

Final decomposition
$$
\boxed{1800 = 2^{3} \, 3^{\color{blue}{2}} \, 5^{\color{red}{2}}}
$$


Start with a list of primes:
$$
 p =\left\{
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,
67, 71
\right\}
$$

If you were not given the $2^{3}$ at the beginning, you would start with $p=2$
$$
 \underbrace{\frac{1800}{2} = 900}_{1} 
\qquad \Rightarrow \qquad
 \underbrace{\frac{900}{2} = 450}_{2}
\qquad \Rightarrow \qquad
 \underbrace{\frac{450}{2} = 225}_{\color{green}{3}}
\qquad \Rightarrow \qquad
 \frac{225}{2} \notin \mathbb{Z}
$$
The $\color{green}{3}$ successful divisions reveal $p^{\color{green}{3}}=2^{\color{green}{3}}$ is a factor.
