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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##