How do you solve for x in # 1/log_3x + 1/log _27 x = 4#?

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Answer 1 · 2018-07-28T03:30:04

#x=3#

Just a simple thing to note: #1/(log_a(b))=log_b(a)#

Therefore,

#1/(log_3(x))+1/(log_27(x))=4#

#=>log_x(3)+log_x(27)=4#

Using the fact that #log_a(b)+log_a(c)=log_a(bc)#,

#=>log_x(3*27)=4#

#=>log_x(81)=4#

Since #log_a(b)=c# is just #a^c=b,#

#x^4=81# We can solve this equation!

#=>x=root[4] (81)#

#=>x=3# or #x=-3#
For now, you can just ignore the negative answer.

#x=3#

Please ask if you are confused or curious about my answer!

Answer 2 · 2018-07-28T03:33:30

#x=3#

Given equation:

#1/\log_3x+1/\log_{27}x=4#

#1/\log_3x+1/\log_{3^3}x=4#

#1/\log_3x+1/{1/3\log_{3}x}=4\quad (\because \ \log_{a^n}b=1/n\log_a b)#

#1/\log_3x+3/{\log_{3}x}=4#

#4/{\log_{3}x}=4#

#\log_{3}x=4/4#

#\log_3x=1#

#x=3#