[MrBlackwood]

The following has caught me out;

2^x = 512

I have no idea where to start, but I think the use of logs is needed?

Could someone point me in the right direction?

Rep for useful response!

[Intriguing Alias]

(Original post by MrBlackwood)
The following has caught me out;

2^x = 512

I have no idea where to start, but I think the use of logs is needed?

Could someone point me in the right direction?

Rep for useful response!

Take log base 2 of both sides (or if your calculator doesn't support that - just take regular logs of both sides).

This is one of the most basic logarithm questions - have you studied logs properly yet?

[MrBlackwood]

(Original post by hassi94)
Take log base 2 of both sides (or if your calculator doesn't support that - just take regular logs of both sides).

This is one of the most basic logarithm questions - have you studied logs properly yet?

Thank you!!

And no, haha. We were taught the rules in 3 minutes and expected to be experts with no further teaching

[Josephdude]

Or if you have a calculator...you can simply divide 512 by 2 for 9 times...

hence you get 2^9 = 512...

(but taking log I think is the better option)

[Sora]

Loga^x = xLoga

Hopefully this is a nudge in the right direction :P

[Hasufel]

could also take natural logs...

[iCiaran]

(Original post by Hasufel)
could also take natural logs...

The base is irrelevant as long as it is the same on both sides, just happens that in this case base 2 is convenient for the 512...