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Bedmas for Log equations

Hi there

I was wondering how does the order of operations apply to log equations.

Thanks

Logarithms are essentially the inverse of exponents so they lie in the same step when using bedmas. You apply the same rules when combining logarithms (with the same base) as you use for exponential functions:

(x^a)*(x^b) = x^(a+b)
logx(a) + logx(b) = logx(ab)

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Hi there, thank you for your response.
So just to clarify, if there were an equation that we had to simplify such as the one below, we would perform the addition part of the equation first as per bedmas rules.

logx(a) - logx(b) + logx(c)

That is a very good question. The fact is subtraction and addition are the same order of operations (subtraction is addition of a negative number), none of them goes before another, you just apply them from left to right as they are written. So in your example

logx(a) - logx(b) + logx(c) = logx (a/b * c)

So we come to the next idea - multiplication and division also have the same order of operations as division is reciprocal mulitplication ( a divided by b = a*1/b).

So in your example you do subtraction, then addition:

logx(a) - logx(b) + logx(c) = logx(a) + (- logx(b)) + logx(c) = logx (a/b * c) = logx (a*1/b * c)

I hope that clears things.

Just be careful with BEDMAS it isn’t exactly in that particular order. Brackets first, followed by exponents, division and multiplication are of the same order (work left to right), then lastly addition and subtraction also have the same order (where again you work left to right). This is a commonly miss-understood concept.