How to work out value of $a$ and $b$ in exponents?

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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.

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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.