#2^x=64#
Just in case your teacher wants you to solve this the fancy way...
#log2^x=log64color(white)(aaa)#Take the log of both sides
#xlog2=log64color(white)(aaa)#Use the log rule #logx^a=alogx#
#(xlog2)/log2=log64/log2color(white)(aaa)#Divide both sides by log2
#x=6#
Or, if you're not allowed to use a calculator...
#log_2 2^x=log_2 64#
#xlog_2 2=log_2 64#
#(xlog_2 2)/log_2 2=log_2 64/log_2 2#
#log_color(red)2 color(violet)(64) = color(blue)6# because the answer to a log problem is the exponent #color(blue)6# that will make the base #color(red)2# equal the argument #color(violet)(64)#. In other words, #color(red)2^color(blue)6=color(violet)(64)#. Similarly, #log_color(red)2 color(violet)(2) =color(blue)1# because #color(red)2^color(blue)1=color(violet)2#. But if you knew that, you could solve this problem without all the fancy math!
#x=6/1#
#x=6#
