How do you solve #\ln x = 2\ln 3#?

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Answer 1 · 2018-05-22T19:39:28

#x = 9#

use the log law: #a log x = log x^a#
(note that #ln x = log_e x#, so the law can also be used for #ln# expressions)

here, #a ln x = 2 ln 3#
#a = 2, x = 3#

#ln x^a#, then, #= ln 3^2#
#ln 3^2 = ln 9#

if #ln x = ln 9#, then #x = 9#

Answer 2 · 2018-05-22T19:46:57

#x=9#

#"using the "color(blue)"laws of logarithms"#

#•color(white)(x)logx^nhArrnlogx#

#•color(white)(x)logx=logyrArrx=y#

#lnx=ln3^2=ln9rArrx=9#