# advanced algebra

we are working with logarithms and i don't understand them at all! I get the basic content but then it gets a little more advanced, and I get completely confused.

There's one in particular that I have NO idea how to do. Help please?!

Simplify:
3 ^ [log(base 9)2 + log(base 9)4]

Thank you.. =]

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1. when you add logs of the same base, you are multiplying
like
log A + log B = log (A*B)
so we have really

3^log9(8)
--------------------------
ONE WAY --If you know log rules
logb(M) = logc(M)/logc(b)
so
log9(8) = log3(8)/log3(9)
but
log3 (9) = log3 (3)^2 = 2*1 = 2
so
log9(8) = log 3(8)/2 = (1/2)log3(8)
so now back to the problem
3^log9(8) = 3^(1/2) log3(8)
= (3^log3(8))^1/2
but b^logb(x) = x
so this is 8^(1/2)
= sqrt(8) = 2 sqrt(2)
----------------------------
Another more basic way:
Now a basic fact of logs is that base^log x = x
so 9^log9(8) = 8

so (3^2)^log9(8) = 8

so 3^2log9(8) = 8

so [3^log9(8)]^2 = 8

so 3^log9(8) = sqrt 8 = 2 sqrt 2

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