Solving Logarithmic Equations: Find log[base 2]log[base 3]a = 2

[AzureNight]

I have a question I need to do which I'm stuck on:

find a

log[base 2](log[base 3]a) = 2

How would I do that? And sorry about the crude representation of the base, but I'm not sure how to do a small 2 or 3 on the computer.

Edit: oh damn, wrong forum. I guess it has to be moved to the appropriate one for homework questions.

 

[d_leet]

What have you tried so far? What is the definition of a logarithm?

 

[kesh]

suggest you think about the inverse of logs of an integer base and how easily they can be worked out

 

[Gib Z]

Ok well. log[base a] x = y then a^y= x, by definition.

Then in this case, 2^2 = log[base 3] a
log[base3]a = 4

Indentity- log[base x] y = (ln y)/(ln x) Hopefully uve learned this.

ln a = 4 ln 3

make both sides the exponents of e.
a= e^(4 ln3 )

Therefore, a = 81.

 

[kesh]

Ok well. log[base a] x = y then a^y= x, by definition.

Then in this case, 2^2 = log[base 3] a
log[base3]a = 4

Indentity- log[base x] y = (ln y)/(ln x) Hopefully uve learned this.

ln a = 4 ln 3

make both sides the exponents of e.
a= e^(4 ln3 )

Therefore, a = 81.

no need for e or your identity, just use your 1st line defn again

 

[Gib Z]

o yea that should have been obvious, my bad