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# Rewrite log2(x)+3log2(y)–3log2(z) into a single expression?

I get log2(x+y^3-z^3) but my quiz keeps telling me its wrong... Is there something I'm missing?

### 5 Answers

- 8 years agoFavorite Answer
Your not following the laws of logs, your just combining the arguments.

Recall log (x) + log (y) = log (xy)

Also: log (x) - log (y) = log (x/y)

So for your equation do one then the other.

Rewrite: first parts that are added

Log2(xy^3) - log2 (z^3)

Log2 (xy^3/z^3)

- 8 years ago
Your answer is wrong, and the other one. This is you want to do, log2x+log(2y)^3-log(2z)^3, then you will get log{2x(2y)^3/(2z)^3}. You can even simplify more by doing log{2xy^3/z^3}. I cancel 2^3 from top and bottom.

- Anonymous4 years ago
a million + cos(4x) Use the double attitude formula: cos(2u) = cos^2(u) - sin^2(u), with u=2x a million + cos^2(2x) - sin^2(2x) (a million - sin^2(2x)) + cos^2(2x) 2 cos^2(2x) that's purely approximately what you have, different than for a "2x" interior the arg to cos

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- 8 years ago
try searching for the properties of logarithms!....

log(mn)=log(m)+log(n)

so......{log[(2x).(2y)^3]} /log {(2z)^3}....as log m - log n=log(m/n)