Question

Can someone explain why the answer is 6 (problem having to do with logarithms)? (pic in the comments section)

Answer 1

So that equation is \[\log_{\sqrt{3}} 27\]

So that will be \[\log_{\sqrt{3}} 27=6\]

There are two rules, well not really rules, but equations you need to know.

\[\log_{b} y=x\] is equivalent to \[b ^{x}=y\].

Answer 2

Try to plug your numbers in the b^x=y equation and see if you notice something.

Answer 3

So I have \[\sqrt{3}^{6}=27\]

Answer 4

I know the square root symbol is the same as saying 3 to the 1/2 power. But, I'm not really sure what to make of that.

Answer 5

Ok. so the square root, dont think of it being too complicated. Think of it being times by itself 6 times. for example: \[\sqrt{3}\times \sqrt{3}=\sqrt{9}=3\]

Answer 6

Do you get up to this part?

Answer 7

And so
\[\sqrt{3}\times \sqrt{3}\times \sqrt{3}\times \sqrt{3}\times \sqrt{3}\times \sqrt{3}=3\times3\times3\]

Answer 8

You can do it easier in your head, and say Oh. Its just squareroot of 9 which is 3 to the third power.

Answer 9

Therefore 3x3x3=27

Answer 10

And that proves the log equation is equal to 6.

Answer 11

Oh, okay! Got it, thank you!

Answer 12

No problem. Glad I helped.