precalculus
posted by marcus .
x= 3^log_3_^8 ? The log being in the exponent is throwing me off.

I don't get the questionwhat are the dashes for?

No reason, basically the equation is supposed to read 3^log base 3^8?

recall the definition of logarithm:
b^log_b(n) = n
3^log_3(8) = 8
log_3(8) is the exponent of 3 which produces 8
log_10(100) = 2 because 10^2 = 100