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How do you solve ##ln(x^2)=16##?
I found: ##x=e^8=2,980.95##
We can use the property of the logs that allows you to take out the exponent of the argument and place it as multiplier in front of the log to get:
##2ln(x)=16##
rearrange:
##ln(x)=16/2=8##
use the definition of log to get:
##x=e^8=2,980.95##