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I have been doing some fairly simple math for fun lately to see what I can remember. Other than my other question that I had about another of my solutions (i.e. this is not a duplicate/already has answers somewhere else), I'm a bit iffy about my solution to another one of my questions that I gave myself, and I want to check if my solution is correct.
The question: Solve for y: $\log_5(3y)+\log_5(9)=\log_5(405)$
How I solved it:
$$\log_5(3y)+\log_5(9)=\log_5(405)$$ $$\iff\log_5((9)(3y))=\log_5(405)$$ $$\iff\log_5(27y)=\log_5(405)$$ $$\implies27y=405$$ $$\implies\frac{\cancel{27}y}{\cancel{27}}=\frac{\cancel{27}\cdot 15}{\cancel{27}}$$ $$\iff y=15$$ My question:
Did I solve this correctly, and if not, what methods could I use to attain the correct answer?
There's probably no need for an answer, but I'll write one anyway. The solution is right and the steps used are all correct, apart from some minor matters of notation. The only way you could have done it faster, maybe, was simplifying by $3$ at the beginning, and isolate the $y$ on the left side by subtraction of $\log_5(9)$, so the property of logarithms and exponentiation gives the result immediately.